| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx \), we can use the method of contour integration in complex analysis. However, a simpler approach for this specific integral is to use trigonometric substitution. Let's use the substitution \( x = \tan(\theta) \), which implies \( dx = \sec^2(\theta)d\theta \). Also, note that \( x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta) \). Substituting these into the integral, we get: \[ I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(\sec^2(\theta))^2} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) d\theta. \] Now, we use the double angle formula for cosine, \( \cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta)) \), to simplify the integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(1 + \cos(2\theta)) d\theta = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) d\theta. \] This integral is straightforward to compute: \[ I = \frac{1}{2} \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) - \left( -\frac{\pi}{2} + \frac{1}{2}\sin(-\pi) \right) \right) = \frac{\pi}{2}. \] Therefore, the value of the integral is \( \boxed{I = \frac{\pi}{2}} \). |
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| deepseekr1 | To compute the integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx \), we can use both real analysis and complex analysis methods. **Real Analysis Approach:** 1. **Substitution:** Let \( x = \tan \theta \), then \( dx = \sec^2 \theta \, d\theta \). When \( x \) ranges from \( -\infty \) to \( \infty \), \( \theta \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). 2. **Transform the Integral:** The integral becomes: \[ \int_{-\pi/2}^{\pi/2} \frac{\sec^2 \theta}{(\tan^2 \theta + 1)^2} \, d\theta = \int_{-\pi/2}^{\pi/2} \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta = \int_{-\pi/2}^{\pi/2} \cos^2 \theta \, d\theta \] 3. **Power-Reduction Formula:** Using \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \): \[ \int_{-\pi/2}^{\pi/2} \cos^2 \theta \, d\theta = \frac{1}{2} \int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta) \, d\theta \] 4. **Evaluate the Integrals:** \[ \frac{1}{2} \left[ \int_{-\pi/2}^{\pi/2} 1 \, d\theta + \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta \right] = \frac{1}{2} \left[ \pi + 0 \right] = \frac{\pi}{2} \] **Complex Analysis Approach:** 1. **Contour Integral:** Consider the function \( f(z) = \frac{1}{(z^2 + 1)^2} \) with poles at \( z = i \) and \( z = -i \). We use the upper half-plane contour enclosing \( z = i \). 2. **Residue Calculation:** The pole at \( z = i \) is of order 2. Compute the residue: \[ \text{Res}(f, i) = \frac{1}{1!} \frac{d}{dz} \left[ (z - i)^2 \frac{1}{(z^2 + 1)^2} \right]_{z = i} = \frac{d}{dz} \left( \frac{1}{(z + i)^2} \right)_{z = i} = -\frac{2}{(2i)^3} = \frac{1}{4i} \] 3. **Residue Theorem:** The integral is \( 2\pi i \) times the residue: \[ 2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2} \] Both methods confirm that the value of the integral is \(\boxed{\dfrac{\pi}{2}}\). |
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| gpto1 | To compute the integral: \[ I = \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^2} \] First, notice that the function is even because \( f(-x) = f(x) \). Therefore, we can write: \[ I = 2 \int_{0}^{\infty} \frac{dx}{(x^2 + 1)^2} \] Our goal is to find the indefinite integral \( \int \frac{dx}{(x^2 + 1)^2} \), which will help us compute the definite integral. **Finding the Indefinite Integral:** Let’s find the antiderivative of \( \frac{1}{(x^2 + 1)^2} \). Observe that: \[ \int \frac{dx}{(x^2 + 1)^2} = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan{x} + C \] **Verification of the Antiderivative:** To verify, we differentiate the right-hand side: \[ \frac{d}{dx} \left( \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan{x} \right) \] Differentiate term by term: 1. Differentiate \( \frac{x}{2(x^2 + 1)} \): \[ \frac{d}{dx} \left( \frac{x}{2(x^2 + 1)} \right) = \frac{(1)(x^2 + 1) - x(2x)}{2(x^2 + 1)^2} = \frac{1 - x^2}{2(x^2 + 1)^2} \] 2. Differentiate \( \frac{1}{2} \arctan{x} \): \[ \frac{d}{dx} \left( \frac{1}{2} \arctan{x} \right) = \frac{1}{2} \cdot \frac{1}{1 + x^2} \] Add the derivatives: \[ \frac{1 - x^2}{2(x^2 + 1)^2} + \frac{1}{2} \cdot \frac{1}{x^2 + 1} = \frac{1 - x^2 + x^2 + 1}{2(x^2 + 1)^2} = \frac{2}{2(x^2 + 1)^2} = \frac{1}{(x^2 + 1)^2} \] So, the antiderivative is correct. **Computing the Definite Integral:** Now, evaluate the antiderivative at the limits from \(-\infty\) to \(\infty\): \[ I = \lim_{b \to \infty} \left( \frac{b}{2(b^2 + 1)} + \frac{1}{2} \arctan{b} \right) - \lim_{a \to -\infty} \left( \frac{a}{2(a^2 + 1)} + \frac{1}{2} \arctan{a} \right) \] As \( b \to \infty \): - \( \frac{b}{2(b^2 + 1)} \to 0 \) because the denominator grows faster than the numerator. - \( \arctan{b} \to \frac{\pi}{2} \). As \( a \to -\infty \): - \( \frac{a}{2(a^2 + 1)} \to 0 \). - \( \arctan{a} \to -\frac{\pi}{2} \). Substitute: \[ I = \left( 0 + \frac{1}{2} \cdot \frac{\pi}{2} \right) - \left( 0 + \frac{1}{2} \cdot \left( -\frac{\pi}{2} \right) \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \] **Answer:** \[ I = \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^2} = \frac{\pi}{2} \] **Final Result:** \(\displaystyle I\;=\;\int_{-\infty}^\infty\frac{dx}{(x^{2}+1)^{2}}\;=\;\frac{\pi}{2}\) |
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| GPT-4o | To evaluate the integral \[ I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx, \] we will use a method involving contour integration in the complex plane. Here's a step-by-step solution: 1. **Complex Function and Contour Choice:** Define the complex function \( f(z) = \frac{1}{(z^2 + 1)^2} \). The poles of this function are the solutions to \( z^2 + 1 = 0 \), which are \( z = \pm i \). Both poles are of order 2 (double poles). 2. **Residue Calculation:** We need to calculate the residues at the poles \( z = i \) and \( z = -i \). For a double pole, the residue at \( z = i \) is given by: \[ \text{Res}(f, i) = \lim_{z \to i} \frac{d}{dz} \left[ (z - i)^2 f(z) \right]. \] First, compute \( (z - i)^2 f(z) \): \[ (z - i)^2 f(z) = \frac{(z - i)^2}{(z^2 + 1)^2}. \] Since \( z^2 + 1 = (z - i)(z + i) \), we have: \[ (z^2 + 1)^2 = (z - i)^2 (z + i)^2. \] Thus, \[ (z - i)^2 f(z) = \frac{(z - i)^2}{(z - i)^2 (z + i)^2} = \frac{1}{(z + i)^2}. \] Now, differentiate \( \frac{1}{(z + i)^2} \) with respect to \( z \): \[ \frac{d}{dz} \left( \frac{1}{(z + i)^2} \right) = -\frac{2}{(z + i)^3}. \] Evaluating this at \( z = i \): \[ \left. -\frac{2}{(z + i)^3} \right|_{z = i} = -\frac{2}{(2i)^3} = -\frac{2}{8i^3} = -\frac{2}{8(-i)} = \frac{1}{4i}. \] 3. **Residue at \( z = -i \):** Similarly, for \( z = -i \): \[ \text{Res}(f, -i) = \lim_{z \to -i} \frac{d}{dz} \left[ (z + i)^2 f(z) \right]. \] Following the same steps: \[ (z + i)^2 f(z) = \frac{(z + i)^2}{(z - i)^2 (z + i)^2} = \frac{1}{(z - i)^2}. \] Differentiate \( \frac{1}{(z - i)^2} \): \[ \frac{d}{dz} \left( \frac{1}{(z - i)^2} \right) = -\frac{2}{(z - i)^3}. \] Evaluating at \( z = -i \): \[ \left. -\frac{2}{(z - i)^3} \right|_{z = -i} = -\frac{2}{(-2i)^3} = -\frac{2}{-8i^3} = -\frac{2}{-8(-i)} = \frac{1}{4i}. \] 4. **Sum of Residues:** The residues at the poles \( i \) and \( -i \) are both \( \frac{1}{4i} \). The sum of the residues is: \[ \text{Res}(f, i) + \text{Res}(f, -i) = \frac{1}{4i} + \frac{1}{4i} = \frac{1}{2i}. \] 5. **Application of Residue Theorem:** According to the residue theorem applied to a semicircular contour that encloses the upper half-plane (or lower half-plane), the integral over the real axis is: \[ \int_{-\infty}^{\infty} f(x) \, dx = 2\pi i \times \left( \text{ |
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| llama3.1-70B-Instruct | To compute the real integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx \), we can use the method of contour integration from complex analysis, specifically the residue theorem. However, since the problem asks for a real integral, we'll also outline a real analysis approach using trigonometric substitution. ### Real Analysis Approach 1. **Trigonometric Substitution**: Let \( x = \tan(\theta) \), which implies \( dx = \sec^2(\theta) d\theta \). The limits of integration change to \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \) because \( \tan(\theta) \) approaches \( \pm\infty \) as \( \theta \) approaches \( \pm\frac{\pi}{2} \). 2. **Substitute into the Integral**: With \( x = \tan(\theta) \), we have \( x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta) \). Thus, the integral becomes: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{\sec^4(\theta)} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) d\theta \] 3. **Evaluate the Integral**: Using the trigonometric identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \), we get: \[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 + \cos(2\theta) d\theta \] \[ I = \frac{1}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \] \[ I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + 0 \right) - \left( -\frac{\pi}{2} + 0 \right) \right] \] \[ I = \frac{\pi}{2} \] ### Complex Analysis Approach For completeness, let's mention the complex analysis approach, which involves integrating around a contour that includes the real axis from \( -R \) to \( R \) and a semicircle \( \Gamma \) of radius \( R \) in the upper half-plane. The integral around the contour is \( 2\pi i \) times the sum of the residues of the poles inside the contour. The poles of \( \frac{1}{(z^2 + 1)^2} \) are at \( z = \pm i \), but only \( z = i \) is inside the contour. The residue at \( z = i \) can be calculated, and as \( R \) approaches infinity, the integral around the semicircle approaches zero, leaving the integral along the real axis as the solution. This approach also yields \( I = \frac{\pi}{2} \). In conclusion, both the real analysis and complex analysis approaches give the same result for the integral \( I \). |
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| llama3.1-405B-Instruct | To compute the real integral $I=\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx$, we can use the method of contour integration. First, we note that the integrand is an even function, so we can write: $$I = 2 \int_0^{\infty} \frac{1}{(x^2 + 1)^2} dx$$ Next, we consider the contour integral: $$\oint_C \frac{1}{(z^2 + 1)^2} dz$$ where $C$ is the contour consisting of the real axis from $-R$ to $R$ and the semicircle $\Gamma$ of radius $R$ in the upper half-plane. Using the residue theorem, we can evaluate the contour integral as: $$\oint_C \frac{1}{(z^2 + 1)^2} dz = 2\pi i \sum_{k=1}^2 \text{Res}\left(\frac{1}{(z^2 + 1)^2}, z_k\right)$$ where $z_1 = i$ and $z_2 = -i$ are the poles of the integrand. Computing the residues, we get: $$\text{Res}\left(\frac{1}{(z^2 + 1)^2}, i\right) = \frac{1}{4i}$$ $$\text{Res}\left(\frac{1}{(z^2 + 1)^2}, -i\right) = -\frac{1}{4i}$$ Therefore, the contour integral is: $$\oint_C \frac{1}{(z^2 + 1)^2} dz = 2\pi i \left(\frac{1}{4i} - \frac{1}{4i}\right) = 0$$ Now, we can write: $$\int_{-R}^R \frac{1}{(x^2 + 1)^2} dx + \int_{\Gamma} \frac{1}{(z^2 + 1)^2} dz = 0$$ As $R \to \infty$, the second integral vanishes, and we are left with: $$\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = 0$$ Therefore, the value of the integral is $I = 0$. |
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| DeepSeek-Coder-V2 | To compute the real integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx \), we can use the method of contour integration in complex analysis. Here is the step-by-step solution: 1. **Define the Contour**: Consider a contour \( C \) that consists of the real axis from \( -R \) to \( R \) and the semicircle in the upper half-plane from \( R \) to \( -R \), where \( R \) is a large positive number. The contour integral is: \[ \oint_C \frac{1}{(z^2 + 1)^2} \, dz \] 2. **Analyze the Poles**: The function \( \frac{1}{(z^2 + 1)^2} \) has poles at \( z = i \) and \( z = -i \). Since we are considering the upper half-plane, we only need to consider the pole at \( z = i \). 3. **Residue Calculation**: To find the residue at \( z = i \), we use the fact that \( z = i \) is a double pole. The residue \( \text{Res} \left( \frac{1}{(z^2 + 1)^2}, i \right) \) can be calculated as follows: \[ \text{Res} \left( \frac{1}{(z^2 + 1)^2}, i \right) = \lim_{z \to i} \frac{d}{dz} \left( (z - i)^2 \frac{1}{(z^2 + 1)^2} \right) \] First, simplify the function: \[ \frac{1}{(z^2 + 1)^2} = \frac{1}{(z + i)^2 (z - i)^2} \] Then, differentiate: \[ \frac{d}{dz} \left( \frac{1}{(z + i)^2} \right) = -\frac{2}{(z + i)^3} \] Therefore, \[ \text{Res} \left( \frac{1}{(z^2 + 1)^2}, i \right) = \lim_{z \to i} -\frac{2}{(z + i)^3} = -\frac{2}{(2i)^3} = -\frac{2}{-8i} = \frac{1}{4i} \] 4. **Contour Integral**: By the residue theorem, the contour integral over \( C \) is: \[ \oint_C \frac{1}{(z^2 + 1)^2} \, dz = 2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2} \] 5. **Estimate the Semi-Circle Contribution**: As \( R \to \infty \), the integral over the semicircle vanishes. This can be shown by estimating the magnitude of the integrand on the semicircle: \[ \left| \int_{\text{semicircle}} \frac{1}{(z^2 + 1)^2} \, dz \right| \leq \int_{\text{semicircle}} \frac{1}{|z^2 + 1|^2} \, |dz| \leq \int_{\text{semicircle}} \frac{1}{(R^2 - 1)^2} \, |dz| \to 0 \text{ as } R \to \infty \] 6. **Combine Results**: Therefore, the integral over the real axis is equal to the contour integral: \[ \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx = \frac{\pi}{2} \] Thus, the value of the integral \( I \) is: \[ I = \frac{\pi}{2} \] |
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| Qwen2-72B-Instruct | To solve the given integral $I=\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx$, we can use the method of contour integration in complex analysis. However, a simpler approach for this specific integral is to use trigonometric substitution. Let's use the substitution $x = \tan(\theta)$, which implies $dx = \sec^2(\theta)d\theta$. Also, note that $x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta)$. Substituting these into the integral, we get: $$I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(\sec^2(\theta))^2} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) d\theta.$$ Now, we use the double angle formula for cosine, $\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))$, to simplify the integral: $$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(1 + \cos(2\theta)) d\theta = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) d\theta.$$ This integral is straightforward to compute: $$I = \frac{1}{2} \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) - \left( -\frac{\pi}{2} + \frac{1}{2}\sin(-\pi) \right) \right) = \frac{\pi}{2}.$$ Therefore, the value of the integral is $\boxed{I = \frac{\pi}{2}}$. |
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| baichuan4 | To compute the real integral $$I=\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx$$, we can use the residue theorem from complex analysis. This involves considering the complex integral along a semicircular contour in the upper half-plane and taking the limit as the radius of the semicircle goes to infinity. First, let's find the poles of the function $$f(z) = \frac{1}{(z^2 + 1)^2}$$. The poles are the solutions to the equation $$z^2 + 1 = 0$$, which are at $$z = i$$ and $$z = -i$$. Since the degree of the denominator is twice that of the numerator, these are second-order poles. Next, we need to calculate the residues at these poles. For a second-order pole at $$z_0$$, the residue is given by $$\frac{d}{dz} \left[ (z-z_0)^2 f(z) \right]_{z=z_0}$$. For the pole at $$z_0 = i$$, we have: $$ \text{Res}(f, i) = \left. \frac{d}{dz} \left[ (z-i)^2 \frac{1}{(z^2 + 1)^2} \right] \right|_{z=i} = \left. \frac{d}{dz} \left[ \frac{(z-i)^2}{(z^2 + 1)^2} \right] \right|_{z=i} $$ After differentiating and evaluating at $$z = i$$, we get: $$ \text{Res}(f, i) = \frac{1}{4i} $$ Now, consider the complex integral $$\oint_C \frac{1}{(z^2 + 1)^2} dz$$ where $$C$$ is a semicircular contour in the upper half-plane with radius $$R$$ and center at the origin. By the residue theorem, this integral is equal to $$2\pi i$$ times the sum of the residues of the function inside the contour. Since there is only one pole (at $$z = i$$) inside the contour, we have: $$ \oint_C \frac{1}{(z^2 + 1)^2} dz = 2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2} $$ As $$R$$ goes to infinity, the contribution from the semicircular arc goes to zero because the integrand decays sufficiently fast (as $$1/z^4$$) for large $$z$$. Therefore, in the limit, the integral along the real axis from $$-\infty$$ to $$\infty$$ is equal to the integral around the contour $$C$$. Thus, we have: $$ I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = \frac{\pi}{2} $$ This is the value of the real integral. |
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| doubao | Let's use the residue theorem to evaluate this integral. Consider the function $f(z) = \frac{1}{(z^2+1)^2}$. The poles of $f(z)$ are $z = \pm i$. We need to find the residues at these poles. For $z = i$, we have: \[ \begin{align*} \lim_{z \to i} \frac{d}{dz}[(z-i)^2 f(z)] &= \lim_{z \to i} \frac{d}{dz}[\frac{(z-i)^2}{(z^2+1)^2}]\\ &= \lim_{z \to i} \frac{2(z-i)(z^2+1)^2 - 4z(z-i)^2(z^2+1)}{(z^2+1)^4}\\ &= \lim_{z \to i} \frac{2(z-i)(z^2+1) - 4z(z-i)^2}{(z^2+1)^3}\\ &= \frac{2(i-i)(i^2+1) - 4i(i-i)^2}{(i^2+1)^3}\\ &= \frac{-4i}{(2i)^3} = \frac{1}{4i} \end{align*} \] So the residue at $z = i$ is $\frac{1}{4i}$. By symmetry, the residue at $z = -i$ is also $\frac{1}{4i}$. Now, we integrate $f(z)$ along a semicircle in the upper half-plane with radius $R$ and the real axis. As $R \to \infty$, the integral along the semicircle goes to 0. Then by the residue theorem: \[ \int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2}dx = 2\pi i \left(\frac{1}{4i} + \frac{1}{4i}\right) = \frac{\pi}{2} \] |
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| minimax | To compute the integral $I=\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx$, we can use the method of residues from complex analysis. This method allows us to evaluate real integrals by extending them to the complex plane and using the residue theorem. The integrand $\frac{1}{(x^2 + 1)^2}$ has poles where the denominator is zero, which occurs at $x = i$ and $x = -i$. Since we are integrating along the real axis, we are interested in the pole that lies in the upper half of the complex plane, which is $x = i$. We will consider a semicircular contour in the upper half-plane that goes from $-R$ to $R$ along the real axis and then back to $-R$ through a semicircle of radius $R$ in the upper half-plane. The contour integral over this closed path is given by: $$\oint_C \frac{1}{(z^2 + 1)^2} dz$$ where $C$ is the closed contour consisting of the real axis from $-R$ to $R$ and the semicircular path $C_R$ in the upper half-plane. By the residue theorem, the contour integral is $2\pi i$ times the sum of the residues of the poles inside the contour. Since the only pole inside the contour is at $z = i$, we need to find the residue of $\frac{1}{(z^2 + 1)^2}$ at $z = i$. The residue of a function $f(z)$ at a simple pole $z_0$ is given by: $$\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)$$ However, since our pole is not simple (it's a second-order pole), we need to use a different formula to find the residue. For a pole of order $n$, the residue can be found using: $$\text{Res}(f, z_0) = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}}\left((z - z_0)^n f(z)\right)$$ In our case, $n = 2$, so we need to compute the first derivative of $(z - i)^2 \frac{1}{(z^2 + 1)^2}$ and then take the limit as $z$ approaches $i$. Let's compute the derivative: $$\frac{d}{dz}\left((z - i)^2 \frac{1}{(z^2 + 1)^2}\right) = \frac{d}{dz}\left(\frac{(z - i)^2}{(z^2 + 1)^2}\right)$$ Using the product rule and the chain rule, we get: $$\frac{d}{dz}\left(\frac{(z - i)^2}{(z^2 + 1)^2}\right) = \frac{2(z - i)(z^2 + 1)^2 - (z - i)^2 \cdot 2(z^2 + 1) \cdot 2z}{(z^2 + 1)^4}$$ Simplify the expression: $$= \frac{2(z - i)(z^2 + 1) - 4z(z - i)^2}{(z^2 + 1)^3}$$ Now we need to evaluate this expression as $z$ approaches $i$: $$\lim_{z \to i} \frac{2(z - i)(z^2 + 1) - 4z(z - i)^2}{(z^2 + 1)^3}$$ Plugging in $z = i$: $$= \frac{2(i - i)((i)^2 + 1) - 4i(i - i)^2}{((i)^2 + 1)^3}$$ $$= \frac{0 - 0}{(-1 + 1)^3}$$ $$= \frac{0}{0}$$ This is an indeterminate form, so we need to simplify further or use a different approach. However, we can notice that the numerator simplifies to $-4z(z - i)^2$ and since $z = i$ is a root of $z^2 + 1$, the numerator becomes $-4i(i - i)^2 = 0$. Thus, the residue is actually zero. This means that the integral over the closed contour is zero, and we can use the fact that the integral over the semicircular path $C_R$ in the upper half-plane goes to zero as $R$ goes to infinity (by Jordan's lemma, since the integrand goes to zero faster than $1/R$ as $R \to \infty$). Therefore, the integral over the real line is equal to the integral over the closed contour, which is zero. So the value of the integral $I$ is: $$I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = 0$$ However, this is incorrect, as we have made a mistake in the residue calculation. Let's correct this. The correct residue calculation for the pole at $z = i$ is as follows: $$\text{Res}(f, i) = \lim_{z \to i} \frac{d}{dz}\left((z - i)^2 \frac{1}{(z^2 + 1)^2}\right)$$ We already have the derivative of the function, so we need to evaluate it at $z = i$: $$\text{Res}(f, i) = \left.\frac{2(z - i)(z^2 + 1) - 4z(z - i)^2}{(z^2 + 1)^3}\right|_{z=i}$$ $$= \frac{2(i - i)((i)^2 + 1) - 4i(i - i)^2}{((i)^2 + 1)^3}$$ $$= \frac{0 - 0}{(-1 + 1)^3}$$ $$= \frac{0}{0}$$ Again, we have an indeterminate form. To resolve this, we need to simplify the expression further. Let's go back to the derivative and plug in $z = i$: $$\frac{d}{dz}\left(\frac{(z - i)^2}{(z^2 + 1)^2}\right) = \frac{2(z - i)(z^2 + 1) - (z - i)^2 \cdot 2(z^2 + 1) \cdot 2z}{(z^2 + 1)^4}$$ $$= \frac{2(z - i)(z^2 + 1) - 4z(z - i)^2}{(z^2 + 1)^4}$$ Now, let's plug in $z = i$: $$= \frac{2(i - i)((i)^2 + 1) - 4i(i - i)^2}{((i)^2 + 1)^4}$$ $$= \frac{0 - 0}{(-1 + 1)^4}$$ $$= \frac{0}{0}$$ This is still an indeterminate form, and we need to simplify the expression further. Let's look at the numerator again: $$2(z - i)(z^2 + 1) - 4z(z - i)^2$$ When $z = i$, the numerator simplifies to: $$2(i - i)((i)^2 + 1) - 4i(i - i)^2$$ $$= 2(0)(-1 + 1) - 4i(0)^2$$ $$= 0$$ This is incorrect, as we are not properly simplifying the expression. Let's try a different approach to find the residue. The residue at $z = i$ can be found by differentiating the denominator and then evaluating the limit: $$\text{Res}(f, i) = \lim_{z \to i} \frac{d}{dz}\left(\frac{1}{(z^2 + 1)^2}\right)$$ The derivative of the denominator $(z^2 + 1)^2$ is: $$\frac{d}{dz} (z^2 + 1)^2 = 2(z^2 + 1) \cdot 2z = 4z(z^2 + 1)$$ Now we take the derivative of the entire function: $$\frac{d}{dz} \left(\frac{1}{(z^2 + 1)^2}\right) = \frac{-4z(z^2 + 1)}{(z^2 + 1)^4}$$ $$= \frac{-4z}{(z^2 + 1)^3}$$ Now we evaluate the limit as $z$ approaches $i$: $$\text{Res}(f, i) = \lim_{z \to i} \frac{-4z}{(z^2 + 1)^3}$$ $$= \frac{-4i}{(i^2 + 1)^3}$$ $$= \frac{-4i}{(-1 + 1 |
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| kimi | To solve the integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx \), we can use a method known as contour integration, which is a technique from complex analysis. This method is particularly useful for evaluating real integrals that are difficult to solve using elementary calculus techniques. Here's a step-by-step solution using contour integration: 1. **Identify the function**: The function we are integrating is \( f(z) = \frac{1}{(z^2 + 1)^2} \). 2. **Choose a contour**: We will use a semicircle in the upper half-plane as our contour because the function has no singularities in the upper half-plane. 3. **Parametrize the contour**: The contour consists of a line segment from \(-R\) to \(R\) along the real axis and a semicircle \( C_R \) of radius \( R \) centered at the origin in the upper half-plane. The parameterization for the real axis is \( z = x \) where \( x \) ranges from \(-R\) to \(R\). For the semicircle, we can use \( z = Re^{i\theta} \) where \( \theta \) ranges from \(0\) to \(\pi\). 4. **Evaluate the integral over the contour**: The integral over the contour \( C \) is given by \( \int_C f(z) \, dz \). This can be split into two parts: the integral over the real axis and the integral over the semicircle. 5. **Evaluate the integral over the real axis**: This is the integral we want to solve, \( \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx \). 6. **Evaluate the integral over the semicircle**: As \( R \) approaches infinity, the integral over the semicircle approaches zero because \( f(z) \) is analytic in the upper half-plane and the magnitude of \( f(z) \) goes to zero as \( z \) moves away from the real axis. 7. **Use the residue theorem**: The residue theorem states that for a function with isolated singularities inside a closed contour, the integral over the contour is equal to \( 2\pi i \) times the sum of the residues of the function at its singularities inside the contour. In this case, the function has simple poles at \( z = i \) and \( z = -i \), but only \( z = i \) is inside our contour. 8. **Find the residue at \( z = i \)**: The residue at \( z = i \) can be found using the formula for the residue of a simple pole, \( \lim_{z \to i} (z - i) f(z) \). After some algebra, we find that the residue is \( \frac{1}{4i} \). 9. **Apply the residue theorem**: Since the integral over the semicircle goes to zero as \( R \) approaches infinity, the integral over the contour is equal to \( 2\pi i \) times the residue at \( z = i \). Therefore, \( \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx = 2\pi i \cdot \frac{1}{4i} \). 10. **Simplify the result**: The \( i \) terms cancel out, and we are left with \( \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx = \frac{\pi}{2} \). So the solution to the integral is \( I = \frac{\pi}{2} \). |
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| wenxinyiyan | 首先，我们注意到被积函数$f(x)=\frac{1}{(x^2+1)^2}$是一个偶函数，即$f(-x)=f(x)$。因此，我们可以将积分区间从$(-\infty, \infty)$缩减为$(0, \infty)$，并乘以2： $$I = 2\int_0^\infty \frac{1}{(x^2+1)^2}dx$$ 接下来，我们进行换元，令$x=\tan(t)$，这样$dx = \sec^2(t)dt$。同时，当$x=0$时，$t=0$；当$x\to\infty$时，$t\to\frac{\pi}{2}$。因此，积分变为： $$I = 2\int_0^{\frac{\pi}{2}} \frac{\sec^2(t)}{(\tan^2(t)+1)^2}dt$$ 由于$\tan^2(t)+1 = \sec^2(t)$，我们可以进一步简化被积函数： $$I = 2\int_0^{\frac{\pi}{2}} \frac{\sec^2(t)}{\sec^4(t)}dt = 2\int_0^{\frac{\pi}{2}} \cos^2(t)dt$$ 现在，我们使用三角恒等式$\cos^2(t) = \frac{1+\cos(2t)}{2}$来求解积分： $$I = \int_0^{\frac{\pi}{2}} (1+\cos(2t))dt = \left[ t+\frac{1}{2}\sin(2t) \right]_0^{\frac{\pi}{2}}$$ 计算上述表达式在积分区间的两个端点的值，我们得到： $$I = \left( \frac{\pi}{2}+0 \right) - \left( 0+0 \right) = \frac{\pi}{2}$$ 因此，最终的答案是： $$I = \frac{\pi}{2}$$ |
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| zhipu | To compute the integral \( I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx \), we can use a variety of methods. One common approach is to use complex analysis, specifically the residue theorem. However, we can also solve it using real analysis techniques. Here, I'll present a solution using a trigonometric substitution. ### Step-by-Step Solution: 1. **Trigonometric Substitution:** Let \( x = \tan \theta \). Then \( dx = \sec^2 \theta \, d\theta \). 2. **Transform the Integral:** Substitute \( x = \tan \theta \) into the integral: \[ I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{(\tan^2 \theta + 1)^2} \sec^2 \theta \, d\theta. \] 3. **Simplify the Integrand:** Note that \( \tan^2 \theta + 1 = \sec^2 \theta \). Therefore, the integrand simplifies: \[ \frac{1}{(\tan^2 \theta + 1)^2} \sec^2 \theta = \frac{1}{\sec^4 \theta} \sec^2 \theta = \cos^2 \theta. \] 4. **New Integral Form:** The integral now becomes: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta. \] 5. **Use the Cosine Squared Identity:** Recall the trigonometric identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2}. \] Substitute this into the integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, d\theta. \] 6. **Split the Integral:** Split the integral into two parts: \[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, d\theta + \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2\theta \, d\theta. \] 7. **Evaluate Each Part:** - The first integral is straightforward: \[ \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, d\theta = \frac{1}{2} \left[ \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) = \frac{1}{2} \cdot \pi = \frac{\pi}{2}. \] - The second integral involves the cosine function: \[ \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2\theta \, d\theta. \] The antiderivative of \( \cos 2\theta \) is \( \frac{1}{2} \sin 2\theta \): \[ \frac{1}{2} \left[ \frac{1}{2} \sin 2\theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{4} \left( \sin \pi - \sin (-\pi) \right) = \frac{1}{4} (0 - 0) = 0. \] 8. **Combine the Results:** Adding the two parts together, we get: \[ I = \frac{\pi}{2} + 0 = \frac{\pi}{2}. \] Therefore, the value of the integral is: \[ I = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} \, dx = \frac{\pi}{2}. \] |
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| gpto1 | To compute the integral: \[ I = \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^2} \] First, notice that the function is even because \( f(-x) = f(x) \). Therefore, we can write: \[ I = 2 \int_{0}^{\infty} \frac{dx}{(x^2 + 1)^2} \] Our goal is to find the indefinite integral \( \int \frac{dx}{(x^2 + 1)^2} \), which will help us compute the definite integral. **Finding the Indefinite Integral:** Let’s find the antiderivative of \( \frac{1}{(x^2 + 1)^2} \). Observe that: \[ \int \frac{dx}{(x^2 + 1)^2} = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan{x} + C \] **Verification of the Antiderivative:** To verify, we differentiate the right-hand side: \[ \frac{d}{dx} \left( \frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan{x} \right) \] Differentiate term by term: 1. Differentiate \( \frac{x}{2(x^2 + 1)} \): \[ \frac{d}{dx} \left( \frac{x}{2(x^2 + 1)} \right) = \frac{(1)(x^2 + 1) - x(2x)}{2(x^2 + 1)^2} = \frac{1 - x^2}{2(x^2 + 1)^2} \] 2. Differentiate \( \frac{1}{2} \arctan{x} \): \[ \frac{d}{dx} \left( \frac{1}{2} \arctan{x} \right) = \frac{1}{2} \cdot \frac{1}{1 + x^2} \] Add the derivatives: \[ \frac{1 - x^2}{2(x^2 + 1)^2} + \frac{1}{2} \cdot \frac{1}{x^2 + 1} = \frac{1 - x^2 + x^2 + 1}{2(x^2 + 1)^2} = \frac{2}{2(x^2 + 1)^2} = \frac{1}{(x^2 + 1)^2} \] So, the antiderivative is correct. **Computing the Definite Integral:** Now, evaluate the antiderivative at the limits from \(-\infty\) to \(\infty\): \[ I = \lim_{b \to \infty} \left( \frac{b}{2(b^2 + 1)} + \frac{1}{2} \arctan{b} \right) - \lim_{a \to -\infty} \left( \frac{a}{2(a^2 + 1)} + \frac{1}{2} \arctan{a} \right) \] As \( b \to \infty \): - \( \frac{b}{2(b^2 + 1)} \to 0 \) because the denominator grows faster than the numerator. - \( \arctan{b} \to \frac{\pi}{2} \). As \( a \to -\infty \): - \( \frac{a}{2(a^2 + 1)} \to 0 \). - \( \arctan{a} \to -\frac{\pi}{2} \). Substitute: \[ I = \left( 0 + \frac{1}{2} \cdot \frac{\pi}{2} \right) - \left( 0 + \frac{1}{2} \cdot \left( -\frac{\pi}{2} \right) \right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \] **Answer:** \[ I = \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^2} = \frac{\pi}{2} \] **Final Result:** \(\displaystyle I\;=\;\int_{-\infty}^\infty\frac{dx}{(x^{2}+1)^{2}}\;=\;\frac{\pi}{2}\) |
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