| 模型 | 模型答案 | 正确答案 |
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| 金声玉亮2.0 | The integral you've given is an interesting one, involving trigonometric functions and the Cauchy principal value due to the singularity at \(t=0\). This type of integral can be solved using contour integration in the complex plane, or by recognizing it as a special case of a known integral. The integral can be rewritten as follows: \[ \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} dt \] To solve this, we can use the property of the Dirichlet integral, which states that: \[ \int_{-\infty}^{+\infty} \frac{\sin(ax)}{x} dx = \pi \text{sgn}(a) \] Where \(\text{sgn}(a)\) is the sign function of \(a\). We can rewrite the integral using the product-to-sum trigonometric identity: \[ \sin(3t) \sin(t/\pi) = \frac{1}{2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right] \] So the integral becomes: \[ \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right)}{t^2} dt \] Now, we can differentiate under the integral sign to get a form that is similar to the Dirichlet integral. This involves integrating the function under the integral sign with respect to a parameter, then differentiating the resulting integral with respect to that parameter. However, for this specific case, we can notice that the integrand is the derivative of a function of the form \(\frac{\sin(x)}{x}\) with respect to \(x\), evaluated at \(x = 3t - \frac{t}{\pi}\) and \(x = 3t + \frac{t}{\pi}\). The integral can then be evaluated by considering the derivative of the Dirichlet integral: \[ \int_{-\infty}^{+\infty} \frac{\sin(ax)}{x} dx = \pi \text{sgn}(a) \] Differentiating both sides with respect to \(a\) gives: \[ \int_{-\infty}^{+\infty} \frac{\cos(ax)}{x} dx = \frac{d}{da} \left( \pi \text{sgn}(a) \right) \] Since the derivative of the sign function is a Dirac delta function, we have: \[ \int_{-\infty}^{+\infty} \frac{\cos(ax)}{x} dx = \pi \delta(a) \] The original integral can be seen as a linear combination of these derivatives, evaluated at \(a = 3 - \frac{1}{\pi}\) and \(a = 3 + \frac{1}{\pi}\). Thus, the value of the integral is: \[ \frac{1}{2} \left[ \pi \delta\left(3 - \frac{1}{\pi}\right) - \pi \delta\left(3 + \frac{1}{\pi}\right) \right] = 0 \] This is because the Dirac delta function is zero everywhere except at its argument, and the arguments here are not zero. Therefore, the value of the integral is: \[ \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} dt = 0 \] This solution requires a more advanced understanding of integral calculus, including the Dirac delta function and differentiation under the integral sign. If you have any further questions or need additional clarification, feel free to ask! |
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| deepseekr1 | To solve the integral \(\int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} \, dt\), we can use integration by parts and trigonometric identities. First, we use the product-to-sum identity for sines: \[ \sin(3t) \sin\left(\frac{t}{\pi}\right) = \frac{1}{2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right] \] This transforms the integral into: \[ \frac{1}{2} \left( \int_{-\infty}^{\infty} \frac{\cos\left((3 - \frac{1}{\pi})t\right)}{t^2} \, dt - \int_{-\infty}^{\infty} \frac{\cos\left((3 + \frac{1}{\pi})t\right)}{t^2} \, dt \right) \] Next, we evaluate the integrals \(\int_{-\infty}^{\infty} \frac{\cos(kt)}{t^2} \, dt\) using integration by parts. Let \(u = \cos(kt)\) and \(dv = \frac{dt}{t^2}\). Then \(du = -k \sin(kt) \, dt\) and \(v = -\frac{1}{t}\). Applying integration by parts: \[ \int_{-\infty}^{\infty} \frac{\cos(kt)}{t^2} \, dt = -k \int_{-\infty}^{\infty} \frac{\sin(kt)}{t} \, dt \] The integral \(\int_{-\infty}^{\infty} \frac{\sin(kt)}{t} \, dt\) is known to be \(\pi \text{ sign}(k)\). Since \(k\) is positive in our case, this simplifies to \(\pi\). Therefore: \[ \int_{-\infty}^{\infty} \frac{\cos(kt)}{t^2} \, dt = -k\pi \] Substituting back into our transformed integral, let \(\alpha = 3 + \frac{1}{\pi}\) and \(\beta = 3 - \frac{1}{\pi}\). The integral becomes: \[ \frac{1}{2} \left( -\beta\pi - (-\alpha\pi) \right) = \frac{1}{2} \left( (\alpha - \beta)\pi \right) \] Calculating \(\alpha - \beta\): \[ \alpha - \beta = \left(3 + \frac{1}{\pi}\right) - \left(3 - \frac{1}{\pi}\right) = \frac{2}{\pi} \] Thus, the integral simplifies to: \[ \frac{1}{2} \left( \frac{2}{\pi} \cdot \pi \right) = \frac{1}{2} \cdot 2 = 1 \] Therefore, the value of the integral is \(\boxed{1}\). |
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| gpto1 | To evaluate the integral: \[ I = \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\dfrac{t}{\pi}\right)}{t^2} \, dt \] we can utilize trigonometric identities and analyze the behavior of the integrand. **Step 1: Use Trigonometric Identities** First, recall the identity for the product of two sine functions: \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] Applying this to the integrand: \[ \sin(3t) \sin\left(\dfrac{t}{\pi}\right) = \frac{1}{2} \left[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) \right] \] **Step 2: Rewrite the Integral** The integral becomes: \[ I = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right)}{t^2} \, dt \] Since both the cosine function and \( \dfrac{1}{t^2} \) are even functions, we can focus on the integral from 0 to \( +\infty \): \[ I = \int_{0}^{\infty} \frac{\cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right)}{t^2} \, dt \] **Step 3: Simplify Using Another Trigonometric Identity** We use the identity for the difference of cosines: \[ \cos A - \cos B = -2 \sin\left(\dfrac{A + B}{2}\right) \sin\left(\dfrac{A - B}{2}\right) \] Compute the averages and differences: \[ \frac{A + B}{2} = \frac{(3t - \dfrac{t}{\pi}) + (3t + \dfrac{t}{\pi})}{2} = 3t \] \[ \frac{A - B}{2} = \frac{(3t - \dfrac{t}{\pi}) - (3t + \dfrac{t}{\pi})}{2} = -\dfrac{t}{\pi} \] Substitute back into the identity: \[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) = -2 \sin(3t) \sin\left(-\dfrac{t}{\pi}\right) \] Because \( \sin(-\theta) = -\sin\theta \), this simplifies to: \[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) = 2 \sin(3t) \sin\left(\dfrac{t}{\pi}\right) \] **Step 4: Substitute Back Into the Integral** Now our integral becomes: \[ I = \int_{0}^{\infty} \frac{2 \sin(3t) \sin\left(\dfrac{t}{\pi}\right)}{t^2} \, dt = 2I \] **Step 5: Solve for I** This equation implies that: \[ I = 2I \] Subtract \( I \) from both sides: \[ 0 = I \] Therefore: \[ I = 0 \] **Conclusion** The value of the integral is zero. **Answer:** 0 |
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| GPT-4o | To evaluate the integral \(\int_{-\infty}^{\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt\), we can use a known result from Fourier transform theory or the properties of specific integrals involving sine functions. First, we can simplify the integrand using a product-to-sum identity for sine functions: \[ \sin(3t) \sin\left(\frac{t}{\pi}\right) = \frac{1}{2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right]. \] Thus, the integral becomes: \[ \int_{-\infty}^{\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt. \] Let us consider each term separately. For the term involving \(\cos\left(3t - \frac{t}{\pi}\right)\): \[ I_1 = \int_{-\infty}^{\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right)}{t^2} \, dt. \] For the term involving \(\cos\left(3t + \frac{t}{\pi}\right)\): \[ I_2 = \int_{-\infty}^{\infty} \frac{\cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt. \] Both integrals \(I_1\) and \(I_2\) are similar in structure. However, integrals of this form are known to converge to zero due to the oscillatory nature of the cosine function and the decay of the \(\frac{1}{t^2}\) term. Thus, \[ I_1 = 0 \quad \text{and} \quad I_2 = 0. \] Therefore, the original integral evaluates to: \[ \int_{-\infty}^{\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt = \frac{1}{2} (I_1 - I_2) = \frac{1}{2} (0 - 0) = 0. \] Hence, the value of the integral is: \[ \boxed{0}. \] |
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| llama3.1-70B-Instruct | A nice integral problem! The integral you provided is a bit challenging, but we can tackle it using some clever techniques. Here's the solution: $$\int_{-\infty}^{+\infty} \sin(3t) \sin(t/\pi) \frac{dt}{t^2}$$ First, we can use the trigonometric identity: $$\sin(a) \sin(b) = \frac{1}{2} \left[ \cos(a-b) - \cos(a+b) \right]$$ Applying this to our integral, we get: $$\int_{-\infty}^{+\infty} \frac{1}{2} \left[ \cos(3t - t/\pi) - \cos(3t + t/\pi) \right] \frac{dt}{t^2}$$ Now, we can split the integral into two separate integrals: $$\frac{1}{2} \int_{-\infty}^{+\infty} \cos(3t - t/\pi) \frac{dt}{t^2} - \frac{1}{2} \int_{-\infty}^{+\infty} \cos(3t + t/\pi) \frac{dt}{t^2}$$ Next, we can use the fact that the integral of a cosine function with a rational argument can be expressed in terms of the sine and cosine functions. Specifically, we have: $$\int \cos(ax) \frac{dx}{x^2} = -\frac{\sin(ax)}{x} - a \int \frac{\sin(ax)}{x} dx$$ Applying this to our two integrals, we get: $$-\frac{1}{2} \left[ \frac{\sin(3t - t/\pi)}{t} + (3 - 1/\pi) \int \frac{\sin(3t - t/\pi)}{t} dt \right] + \frac{1}{2} \left[ \frac{\sin(3t + t/\pi)}{t} - (3 + 1/\pi) \int \frac{\sin(3t + t/\pi)}{t} dt \right]$$ Now, we can use the fact that the integral of a sine function with a rational argument can be expressed in terms of the sine and cosine functions. Specifically, we have: $$\int \frac{\sin(ax)}{x} dx = \frac{\pi}{2} - \arctan\left(\frac{1}{ax}\right)$$ Applying this to our two integrals, we get: $$-\frac{1}{2} \left[ \frac{\sin(3t - t/\pi)}{t} + (3 - 1/\pi) \left( \frac{\pi}{2} - \arctan\left(\frac{1}{(3 - 1/\pi)t}\right) \right) \right] + \frac{1}{2} \left[ \frac{\sin(3t + t/\pi)}{t} - (3 + 1/\pi) \left( \frac{\pi}{2} - \arctan\left(\frac{1}{(3 + 1/\pi)t}\right) \right) \right]$$ Finally, we can evaluate the limits as $t \to \pm \infty$ to get: $$\lim_{t \to \pm \infty} \left[ -\frac{1}{2} \left( \frac{\sin(3t - t/\pi)}{t} + (3 - 1/\pi) \left( \frac{\pi}{2} - \arctan\left(\frac{1}{(3 - 1/\pi)t}\right) \right) \right) + \frac{1}{2} \left( \frac{\sin(3t + t/\pi)}{t} - (3 + 1/\pi) \left( \frac{\pi}{2} - \arctan\left(\frac{1}{(3 + 1/\pi)t}\right) \right) \right) \right]$$ Simplifying, we get: $$\frac{\pi}{2} \left( \frac{1}{3 - 1/\pi} - \frac{1}{3 + 1/\pi} \right) = \frac{\pi}{2} \left( \frac{1}{3 - 1/\pi} - \frac{1}{3 + 1/\pi} ) \right)$$ So, the final answer is: $$\boxed{\frac{\pi}{2} \left( \frac{1}{3 - 1/\pi} - \frac{1}{3 + 1/\pi} \right)}$$ |
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| llama3.1-405B-Instruct | A nice integral! To evaluate this integral, we can use the following steps: 1. Use the product-to-sum identity for sine: $$\sin(3t)\sin(t/\pi) = \frac{1}{2}\left[\cos(3t - t/\pi) - \cos(3t + t/\pi)\right]$$ 2. Substitute this expression into the integral: $$\int_{-\infty}^{\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(3t - t/\pi) - \cos(3t + t/\pi)}{t^2} dt$$ 3. Split the integral into two separate integrals: $$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(3t - t/\pi)}{t^2} dt - \frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(3t + t/\pi)}{t^2} dt$$ 4. Use the fact that the integral of an even function over the entire real line is equal to twice the integral of the function over the positive real line: $$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(3t - t/\pi)}{t^2} dt = \int_{0}^{\infty} \frac{\cos(3t - t/\pi)}{t^2} dt$$ $$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(3t + t/\pi)}{t^2} dt = \int_{0}^{\infty} \frac{\cos(3t + t/\pi)}{t^2} dt$$ 5. Use the following integral identity (which can be proven using contour integration or other methods): $$\int_{0}^{\infty} \frac{\cos(at)}{t^2} dt = \frac{\pi}{2} |a|$$ where $a$ is a real constant. 6. Apply this identity to the two integrals: $$\int_{0}^{\infty} \frac{\cos(3t - t/\pi)}{t^2} dt = \frac{\pi}{2} |3 - 1/\pi|$$ $$\int_{0}^{\infty} \frac{\cos(3t + t/\pi)}{t^2} dt = \frac{\pi}{2} |3 + 1/\pi|$$ 7. Simplify the expressions: $$\frac{\pi}{2} |3 - 1/\pi| = \frac{\pi}{2} \left(3 - \frac{1}{\pi}\right) = \frac{3\pi}{2} - \frac{1}{2}$$ $$\frac{\pi}{2} |3 + 1/\pi| = \frac{\pi}{2} \left(3 + \frac{1}{\pi}\right) = \frac{3\pi}{2} + \frac{1}{2}$$ 8. Combine the results: $$\int_{-\infty}^{\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = \left(\frac{3\pi}{2} - \frac{1}{2}\right) - \left(\frac{3\pi}{2} + \frac{1}{2}\right) = -1$$ Therefore, the value of the integral is: $$\int_{-\infty}^{\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = -1$$ |
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| DeepSeek-Coder-V2 | To solve the integral \(\int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt\), we can use the following steps: 1. **Rewrite the Integrand:** We start by using the product-to-sum identities for sine functions. The product-to-sum formula states: \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] Applying this to our integrand: \[ \sin(3t) \sin\left(\frac{t}{\pi}\right) = \frac{1}{2} \left[\cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right)\right] \] Therefore, the integral becomes: \[ \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt \] 2. **Separate the Integrals:** We can split the integral into two separate integrals: \[ \frac{1}{2} \left[ \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right)}{t^2} \, dt - \int_{-\infty}^{+\infty} \frac{\cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt \right] \] 3. **Evaluate Each Integral:** Notice that the integrands are of the form \(\frac{\cos(at)}{t^2}\). These types of integrals are known to be challenging to evaluate directly due to the singularity at \(t = 0\). However, we can use known results for similar integrals. For the integral \(\int_{-\infty}^{+\infty} \frac{\cos(at)}{t^2} \, dt\), it is known that if \(a \neq 0\), the integral diverges. This is because the integrand behaves poorly near \(t = 0\). 4. **Conclusion:** Since both integrals \(\int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right)}{t^2} \, dt\) and \(\int_{-\infty}^{+\infty} \frac{\cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt\) diverge, the entire expression diverges. Therefore, the value of the integral \(\int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt\) is: \[ \boxed{\text{diverges}} \] |
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| Qwen2-72B-Instruct | The integral you've given is an interesting one, involving trigonometric functions and the Cauchy principal value due to the singularity at \(t=0\). This type of integral can be solved using contour integration in the complex plane, or by recognizing it as a special case of a known integral. The integral can be rewritten as follows: \[ \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} dt \] To solve this, we can use the property of the Dirichlet integral, which states that: \[ \int_{-\infty}^{+\infty} \frac{\sin(ax)}{x} dx = \pi \text{sgn}(a) \] Where \(\text{sgn}(a)\) is the sign function of \(a\). We can rewrite the integral using the product-to-sum trigonometric identity: \[ \sin(3t) \sin(t/\pi) = \frac{1}{2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right] \] So the integral becomes: \[ \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right)}{t^2} dt \] Now, we can differentiate under the integral sign to get a form that is similar to the Dirichlet integral. This involves integrating the function under the integral sign with respect to a parameter, then differentiating the resulting integral with respect to that parameter. However, for this specific case, we can notice that the integrand is the derivative of a function of the form \(\frac{\sin(x)}{x}\) with respect to \(x\), evaluated at \(x = 3t - \frac{t}{\pi}\) and \(x = 3t + \frac{t}{\pi}\). The integral can then be evaluated by considering the derivative of the Dirichlet integral: \[ \int_{-\infty}^{+\infty} \frac{\sin(ax)}{x} dx = \pi \text{sgn}(a) \] Differentiating both sides with respect to \(a\) gives: \[ \int_{-\infty}^{+\infty} \frac{\cos(ax)}{x} dx = \frac{d}{da} \left( \pi \text{sgn}(a) \right) \] Since the derivative of the sign function is a Dirac delta function, we have: \[ \int_{-\infty}^{+\infty} \frac{\cos(ax)}{x} dx = \pi \delta(a) \] The original integral can be seen as a linear combination of these derivatives, evaluated at \(a = 3 - \frac{1}{\pi}\) and \(a = 3 + \frac{1}{\pi}\). Thus, the value of the integral is: \[ \frac{1}{2} \left[ \pi \delta\left(3 - \frac{1}{\pi}\right) - \pi \delta\left(3 + \frac{1}{\pi}\right) \right] = 0 \] This is because the Dirac delta function is zero everywhere except at its argument, and the arguments here are not zero. Therefore, the value of the integral is: \[ \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} dt = 0 \] This solution requires a more advanced understanding of integral calculus, including the Dirac delta function and differentiation under the integral sign. If you have any further questions or need additional clarification, feel free to ask! |
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| baichuan4 | To solve this problem, we can use the fact that the integral of a product of sinusoidal functions can often be simplified using trigonometric identities. Specifically, we'll use the identity: $$\sin{A}\sin{B} = \frac{1}{2}[\cos{(A - B)} - \cos{(A + B)}]$$ Applying this identity to the integrand, we get: $$\int_{-\infty}^{+\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = \int_{-\infty}^{+\infty} \frac{1}{2t^2}[\cos{(3t - t/\pi)} - \cos{(3t + t/\pi)}] dt$$ $$= \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{(3t - t/\pi)}}{t^2} dt - \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{(3t + t/\pi)}}{t^2} dt$$ Now, we can simplify the cosine arguments: $$3t - \frac{t}{\pi} = t\left(3 - \frac{1}{\pi}\right)$$ $$3t + \frac{t}{\pi} = t\left(3 + \frac{1}{\pi}\right)$$ So our integrals become: $$\frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{[t(3 - 1/\pi)]}}{t^2} dt - \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{[t(3 + 1/\pi)]}}{t^2} dt$$ At this point, we can see that these integrals are in the form of the cosine Fourier transform of $1/t^2$. However, the cosine Fourier transform of $1/t^2$ is not a standard result that can be directly looked up. Therefore, we need to find another way to evaluate these integrals. One way to proceed is to recognize that these integrals are even functions over an infinite domain, which means we can use the symmetry property of the cosine function to simplify them. Since the cosine function is even, we can write: $$\frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{[t(3 - 1/\pi)]}}{t^2} dt = \int_{0}^{+\infty} \frac{\cos{[t(3 - 1/\pi)]}}{t^2} dt$$ $$\frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos{[t(3 + 1/\pi)]}}{t^2} dt = \int_{0}^{+\infty} \frac{\cos{[t(3 + 1/\pi)]}}{t^2} dt$$ Now, let's consider the first integral: $$\int_{0}^{+\infty} \frac{\cos{[t(3 - 1/\pi)]}}{t^2} dt$$ We can use integration by parts to evaluate this integral. Let $u = \cos{[t(3 - 1/\pi)]}$ and $dv = 1/t^2 dt$. Then $du = -(3 - 1/\pi)\sin{[t(3 - 1/\pi)]} dt$ and $v = -1/t$. Integration by parts gives: $$\int u dv = uv - \int v du$$ $$= -\frac{\cos{[t(3 - 1/\pi)]}}{t} - \int \frac{-(3 - 1/\pi)\sin{[t(3 - 1/\pi)]}}{t} dt$$ $$= -\frac{\cos{[t(3 - 1/\pi)]}}{t} + (3 - 1/\pi) \int \frac{\sin{[t(3 - 1/\pi)]}}{t} dt$$ The second term can be recognized as the sine Fourier transform of $1/t$, which is a standard result and is equal to $\pi$ for a frequency of $3 - 1/\pi$. Therefore, we have: $$-\frac{\cos{[t(3 - 1/\pi)]}}{t} + (3 - 1/\pi) \pi$$ Taking the limit as $t$ goes from $0$ to $\infty$, the first term goes to $0$ and we are left with: $$(3 - 1/\pi) \pi$$ Similarly, for the second integral: $$\int_{0}^{+\infty} \frac{\cos{[t(3 + 1/\pi)]}}{t^2} dt$$ We can use the same method and find that its value is: $$(3 + 1/\pi) \pi$$ Putting it all together, the original integral is: $$\frac{1}{2}[(3 - 1/\pi) \pi - (3 + 1/\pi) \pi]$$ $$= \frac{1}{2}[3\pi - \pi - 3\pi - \pi]$$ $$= -\pi$$ Therefore, the value of the integral is $-\pi$. |
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| doubao | This integral is a bit challenging to evaluate directly. We can use some properties of Fourier transforms and Parseval's theorem to approach it. Let's define the functions $f(t) = \sin(3t)$ and $g(t) = \frac{\sin(t/\pi)}{t}$. The Fourier transform of $f(t)$ is $F(\omega) = \sqrt{\frac{2\pi}{3}}[\delta(\omega - 3) - \delta(\omega + 3)]$. The Fourier transform of $g(t)$ can be found using standard techniques and is a bit more involved. Then, by Parseval's theorem, the integral we're interested in is related to the integral of the product of the Fourier transforms in the frequency domain. However, carrying out these computations in detail is quite complex and may require advanced techniques and knowledge of Fourier analysis. In conclusion, a complete and detailed solution would involve extensive calculations and possibly the use of specialized tools or software for Fourier transforms. |
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| minimax | To solve the integral $$\int_{-\infty}^{+\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt,$$ we can use the convolution theorem, which states that the Fourier transform of a convolution of two functions is the product of their Fourier transforms. However, we need to be careful with the domain of the functions and the conditions for the convolution theorem to apply. First, let's consider the Fourier transform of the function $f(t) = \frac{\sin(3t)}{t^2}$. The Fourier transform of $f(t)$ is given by $$F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-i\omega t} dt.$$ However, the function $f(t)$ has a singularity at $t = 0$, which means it's not absolutely integrable over the entire real line. To circumvent this, we can consider the principal value of the integral or use a regularization technique. But for the sake of this problem, let's assume we are looking for a distributional solution. The Fourier transform of $\frac{\sin(3t)}{t^2}$ is not straightforward, but we can use the fact that the Fourier transform of $\frac{\sin(at)}{t}$ is $\pi \text{rect}\left(\frac{\omega}{2a}\right)$, where $\text{rect}(x)$ is the rectangular function defined as $$\text{rect}(x) = \begin{cases} 1, & \text{if } |x| \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases}$$ However, the $\frac{1}{t^2}$ factor complicates things, and we would need to use a more advanced technique to find the Fourier transform of $f(t)$, potentially involving distributions and the theory of generalized functions. The function $\sin(t/\pi)$ is periodic with period $2\pi$, and its Fourier series would involve delta functions at the harmonics of the fundamental frequency $1/(2\pi)$. However, the convolution theorem is not directly applicable here because the convolution theorem applies to functions that are absolutely integrable over the real line, and our function has a singularity at $t = 0$. Moreover, the convolution theorem applies to functions that are zero outside a finite interval, which is not the case for $\frac{\sin(3t)}{t^2}$. Instead, we can use the fact that the integral of a product of sine functions over the entire real line can be related to the Dirichlet integral, which is a known result in Fourier analysis. The Dirichlet integral states that $$\int_{-\infty}^{+\infty} \frac{\sin(at)}{t} dt = \pi \text{sgn}(a),$$ where $\text{sgn}(a)$ is the sign function, equal to $-1$ for $a < 0$, $0$ for $a = 0$, and $1$ for $a > 0$. However, the integral we have is not in the form of the Dirichlet integral, and the presence of the $t^2$ in the denominator complicates matters. We would need to use a different approach, potentially involving complex analysis and contour integration. Given the complexity of the integral and the presence of singularities, it would be best to use contour integration in the complex plane to evaluate this integral. This would involve extending the integral to the complex plane and using the residue theorem to compute the integral. However, this is a non-trivial task and requires careful selection of the contour and application of the residue theorem. Unfortunately, without additional context or constraints on the problem, it's not possible to provide a simple closed-form solution to this integral. Advanced techniques from complex analysis and distribution theory would be required to evaluate this integral, and it's likely that the integral does not converge in the classical sense. Therefore, the solution would be in the sense of distributions or generalized functions, which is beyond the scope of a simple answer. |
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| kimi | To solve the integral $$\int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} \, dt,$$ we can use the method of contour integration from complex analysis. This integral is a standard example of an integral that can be evaluated using the residue theorem. First, we rewrite the integrand in terms of complex exponentials using Euler's formula: $$\sin(3t) = \frac{e^{3it} - e^{-3it}}{2i},$$ $$\sin\left(\frac{t}{\pi}\right) = \frac{e^{it/\pi} - e^{-it/\pi}}{2i}.$$ Then, the integrand becomes: $$\frac{\sin(3t) \sin(t/\pi)}{t^2} = \frac{(e^{3it} - e^{-3it})(e^{it/\pi} - e^{-it/\pi})}{4i^2 t^2} = \frac{(e^{3it + it/\pi} - e^{3it - it/\pi} - e^{-3it + it/\pi} + e^{-3it - it/\pi})}{4t^2}.$$ Now, we consider the integral as a contour integral around the real axis with a large semicircle in the upper half-plane. The poles of the integrand are at the zeros of the denominator, which are at $t = 0$ and $t = \pm i$. Since we are interested in the integral from $-\infty$ to $\infty$, we only need to consider the pole at $t = 0$. The residue at $t = 0$ is given by the coefficient of the $\frac{1}{t}$ term in the Laurent series expansion of the integrand. To find this, we can differentiate the numerator with respect to $t$ and then evaluate at $t = 0$: $$\text{Residue} = \lim_{t \to 0} t \frac{d}{dt} \left( \frac{e^{3it + it/\pi} - e^{3it - it/\pi} - e^{-3it + it/\pi} + e^{-3it - it/\pi}}{4t^2} \right).$$ However, this approach can be quite involved. Instead, we can use a known result for integrals of this form. The integral is a special case of the Fourier sine transform of a function, and it is known that: $$\int_{-\infty}^{+\infty} \frac{\sin(ax)}{x^2} \, dx = \frac{\pi}{2} \text{ for } a > 0.$$ In our case, $a = 3$, so the integral evaluates to: $$\int_{-\infty}^{+\infty} \frac{\sin(3t)}{t^2} \, dt = \frac{\pi}{2}.$$ But we also have a factor of $\sin(t/\pi)$, which is an odd function. The product of an odd function and an even function is an odd function, and the integral of an odd function over symmetric limits is zero. Therefore, the integral of the product over the entire real line is zero. So the final answer is: $$\int_{-\infty}^{+\infty} \frac{\sin(3t) \sin(t/\pi)}{t^2} \, dt = 0.$$ |
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| wenxinyiyan | 首先，我们注意到函数$\frac{\sin(3t)\sin(t/\pi)}{t^2}$是偶函数，因为$\sin(3t)$和$\sin(t/\pi)$都是奇函数，而$t^2$是偶函数，奇函数与奇函数的乘积是偶函数，偶函数与偶函数的乘积还是偶函数。因此，我们可以将积分区间从$(-\infty,+\infty)$简化为$[0,+\infty)$，并乘以2： $$\int_{-\infty}^{+\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = 2 \int_{0}^{+\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt$$ 然后，我们可以利用三角函数的乘积化和差公式： $$\sin(3t)\sin(t/\pi) = \frac{1}{2}[\cos(3t-t/\pi)-\cos(3t+t/\pi)]$$ 将上式代入积分中，得到： $$2 \int_{0}^{+\infty} \frac{\sin(3t)\sin(t/\pi)}{t^2} dt = \int_{0}^{+\infty} \frac{\cos(3t-t/\pi)-\cos(3t+t/\pi)}{t^2} dt$$ 接下来，我们可以利用余弦函数的积分性质。考虑到$\cos(3t-t/\pi)$和$\cos(3t+t/\pi)$都是余弦函数的形式，我们可以将其写成$\cos(at+b)$的形式，其中$a$和$b$是常数。然后，我们可以利用以下积分公式： $$\int_{0}^{+\infty} \frac{\cos(at+b)}{t^2} dt = \frac{a\pi}{2}$$ 将上式应用到我们的积分中，得到： $$\int_{0}^{+\infty} \frac{\cos(3t-t/\pi)-\cos(3t+t/\pi)}{t^2} dt = \frac{(3-1/\pi)\pi}{2} - \frac{(3+1/\pi)\pi}{2} = -\pi$$ 因此，原积分的值为$-\pi$。 |
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| zhipu | To solve the integral \[ I = \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\frac{t}{\pi}\right)}{t^2} \, dt, \] we can use a trigonometric identity to simplify the integrand. Recall the product-to-sum identities for sine functions: \[ \sin(A) \sin(B) = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right]. \] Applying this identity to the integrand, we get: \[ \sin(3t) \sin\left(\frac{t}{\pi}\right) = \frac{1}{2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right]. \] Thus, the integral becomes: \[ I = \int_{-\infty}^{+\infty} \frac{1}{2t^2} \left[ \cos\left(3t - \frac{t}{\pi}\right) - \cos\left(3t + \frac{t}{\pi}\right) \right] \, dt. \] This can be split into two separate integrals: \[ I = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right)}{t^2} \, dt - \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt. \] Let's denote these integrals as \( I_1 \) and \( I_2 \): \[ I_1 = \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \frac{t}{\pi}\right)}{t^2} \, dt, \] \[ I_2 = \int_{-\infty}^{+\infty} \frac{\cos\left(3t + \frac{t}{\pi}\right)}{t^2} \, dt. \] Notice that the arguments of the cosine functions in \( I_1 \) and \( I_2 \) are symmetric about \( 3t \). Specifically, if we make the substitution \( u = -t \) in \( I_1 \), we get: \[ I_1 = \int_{-\infty}^{+\infty} \frac{\cos\left(3(-u) - \frac{-u}{\pi}\right)}{(-u)^2} \, (-du) = \int_{-\infty}^{+\infty} \frac{\cos\left(-3u + \frac{u}{\pi}\right)}{u^2} \, du. \] Since \( \cos(-x) = \cos(x) \), this simplifies to: \[ I_1 = \int_{-\infty}^{+\infty} \frac{\cos\left(3u + \frac{u}{\pi}\right)}{u^2} \, du = I_2. \] Therefore, \( I_1 = I_2 \), and the original integral becomes: \[ I = \frac{1}{2} I_1 - \frac{1}{2} I_2 = \frac{1}{2} I_1 - \frac{1}{2} I_1 = 0. \] Thus, the value of the integral is: \[ \boxed{0}. \] |
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| gpto1 | To evaluate the integral: \[ I = \int_{-\infty}^{+\infty} \frac{\sin(3t) \sin\left(\dfrac{t}{\pi}\right)}{t^2} \, dt \] we can utilize trigonometric identities and analyze the behavior of the integrand. **Step 1: Use Trigonometric Identities** First, recall the identity for the product of two sine functions: \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \] Applying this to the integrand: \[ \sin(3t) \sin\left(\dfrac{t}{\pi}\right) = \frac{1}{2} \left[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) \right] \] **Step 2: Rewrite the Integral** The integral becomes: \[ I = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right)}{t^2} \, dt \] Since both the cosine function and \( \dfrac{1}{t^2} \) are even functions, we can focus on the integral from 0 to \( +\infty \): \[ I = \int_{0}^{\infty} \frac{\cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right)}{t^2} \, dt \] **Step 3: Simplify Using Another Trigonometric Identity** We use the identity for the difference of cosines: \[ \cos A - \cos B = -2 \sin\left(\dfrac{A + B}{2}\right) \sin\left(\dfrac{A - B}{2}\right) \] Compute the averages and differences: \[ \frac{A + B}{2} = \frac{(3t - \dfrac{t}{\pi}) + (3t + \dfrac{t}{\pi})}{2} = 3t \] \[ \frac{A - B}{2} = \frac{(3t - \dfrac{t}{\pi}) - (3t + \dfrac{t}{\pi})}{2} = -\dfrac{t}{\pi} \] Substitute back into the identity: \[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) = -2 \sin(3t) \sin\left(-\dfrac{t}{\pi}\right) \] Because \( \sin(-\theta) = -\sin\theta \), this simplifies to: \[ \cos\left(3t - \dfrac{t}{\pi}\right) - \cos\left(3t + \dfrac{t}{\pi}\right) = 2 \sin(3t) \sin\left(\dfrac{t}{\pi}\right) \] **Step 4: Substitute Back Into the Integral** Now our integral becomes: \[ I = \int_{0}^{\infty} \frac{2 \sin(3t) \sin\left(\dfrac{t}{\pi}\right)}{t^2} \, dt = 2I \] **Step 5: Solve for I** This equation implies that: \[ I = 2I \] Subtract \( I \) from both sides: \[ 0 = I \] Therefore: \[ I = 0 \] **Conclusion** The value of the integral is zero. **Answer:** 0 |
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