| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the derivative of \( H(z) \) with respect to \( z \) and then evaluate it at \( z = 1 \), we can use the Leibniz integral rule for differentiating under the integral sign. According to this rule, if we have an integral of the form \[ H(z) = \int_{a(z)}^{b(z)} f(t, z) dt \] the derivative of \( H(z) \) with respect to \( z \) is given by \[ H'(z) = f(b(z), z) b'(z) - f(a(z), z) a'(z) + \int_{a(z)}^{b(z)} \frac{\partial f}{\partial z}(t, z) dt \] In our case, the integral limits are constants (0 and 1), so \( a(z) = 0 \), \( b(z) = 1 \), and \( a'(z) = b'(z) = 0 \). Therefore, the terms involving \( a'(z) \) and \( b'(z) \) vanish. The integrand is \[ f(t, z) = e^{-z^2 t^2} \] and its derivative with respect to \( z \) is \[ \frac{\partial f}{\partial z}(t, z) = -2zt^2 e^{-z^2 t^2} \] Therefore, the derivative of \( H(z) \) with respect to \( z \) is \[ H'(z) = \int_0^1 \frac{\partial f}{\partial z}(t, z) dt = \int_0^1 -2zt^2 e^{-z^2 t^2} dt \] To find \( H'(1) \), we substitute \( z = 1 \) into the expression for \( H'(z) \): \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} dt \] This integral can be solved using integration by parts or recognizing it as related to the error function (erf). However, it does not have a simple closed-form solution in terms of elementary functions. Its value can be approximated numerically or expressed in terms of the error function: \[ H'(1) = -\sqrt{\pi} \frac{d}{dz} \left[ \frac{\text{erf}(zt)}{2z} \right]_{t=0}^{t=1} \Bigg|_{z=1} \] \[ H'(1) = -\sqrt{\pi} \left[ \frac{\text{erf}(1)}{2} - \frac{d}{dz} \left( \frac{\text{erf}(z)}{2z} \right) \Bigg|_{z=1} \right] \] This derivative can be further simplified using the properties of the error function, but for practical purposes, the numerical value can be found using computational tools. |
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| deepseekr1 | To find \( H'(1) \) where \( H(z) = \int_0^1 e^{-z^2 t^2} \, dt \), we use differentiation under the integral sign. First, we compute the partial derivative of the integrand with respect to \( z \): \[ \frac{\partial}{\partial z} e^{-z^2 t^2} = -2z t^2 e^{-z^2 t^2} \] Thus, the derivative \( H'(z) \) is: \[ H'(z) = \int_0^1 -2z t^2 e^{-z^2 t^2} \, dt \] Evaluating this at \( z = 1 \): \[ H'(1) = -2 \int_0^1 t^2 e^{-t^2} \, dt \] We use integration by parts to compute the integral \( \int_0^1 t^2 e^{-t^2} \, dt \). Let \( u = t \) and \( dv = t e^{-t^2} \, dt \). Then \( du = dt \) and \( v = -\frac{1}{2} e^{-t^2} \): \[ \int t^2 e^{-t^2} \, dt = \left. -\frac{t}{2} e^{-t^2} \right|_0^1 + \frac{1}{2} \int_0^1 e^{-t^2} \, dt \] Evaluating the boundary terms: \[ \left. -\frac{t}{2} e^{-t^2} \right|_0^1 = -\frac{1}{2e} + 0 = -\frac{1}{2e} \] The remaining integral is: \[ \frac{1}{2} \int_0^1 e^{-t^2} \, dt = \frac{1}{2} H(1) \] Thus: \[ \int_0^1 t^2 e^{-t^2} \, dt = -\frac{1}{2e} + \frac{1}{2} H(1) \] Substituting back into \( H'(1) \): \[ H'(1) = -2 \left( -\frac{1}{2e} + \frac{1}{2} H(1) \right) = \frac{1}{e} - H(1) \] Since \( H(1) = \int_0^1 e^{-t^2} \, dt \), which can be expressed in terms of the error function \( \text{erf}(1) \): \[ H(1) = \frac{\sqrt{\pi}}{2} \text{erf}(1) \] Therefore: \[ H'(1) = \frac{1}{e} - \frac{\sqrt{\pi}}{2} \text{erf}(1) \] The final answer is: \[ \boxed{\frac{1}{e} - \frac{\sqrt{\pi}}{2} \text{erf}(1)} \] |
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| gpto1 | To find \( H'(1) \), we start by differentiating \( H(z) \) with respect to \( z \): Given: \[ H(z) = \int_0^1 e^{-z^2 t^2} \, dt \] Differentiating under the integral sign (since the limits of integration are constants), we have: \[ H'(z) = \int_0^1 \frac{d}{dz} \left( e^{-z^2 t^2} \right) dt \] Compute the derivative inside the integral: \[ \frac{d}{dz} \left( e^{-z^2 t^2} \right) = e^{-z^2 t^2} \cdot (-2z t^2) = -2z t^2 e^{-z^2 t^2} \] So: \[ H'(z) = \int_0^1 \left( -2z t^2 e^{-z^2 t^2} \right) dt = -2z \int_0^1 t^2 e^{-z^2 t^2} dt \] Evaluating at \( z=1 \): \[ H'(1) = -2 \int_0^1 t^2 e^{-t^2} dt \] Now, let's compute the integral: \[ I = \int_0^1 t^2 e^{-t^2} dt \] Integrate by parts. Let: - \( u = t \), so \( du = dt \) - \( dv = t e^{-t^2} dt \), so \( v = -\frac{1}{2} e^{-t^2} \) (since \( \int t e^{-t^2} dt = -\frac{1}{2} e^{-t^2} + C \)) Using integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ I = u v \Big|_0^1 - \int_0^1 v \, du = \left( t \cdot \left( -\frac{1}{2} e^{-t^2} \right) \right) \Bigg|_0^1 - \int_0^1 \left( -\frac{1}{2} e^{-t^2} \right) dt \] Compute the boundary term: \[ \left( -\frac{1}{2} t e^{-t^2} \right) \Bigg|_0^1 = -\frac{1}{2} \cdot 1 \cdot e^{-1} - (-\frac{1}{2} \cdot 0 \cdot e^{0}) = -\frac{1}{2} e^{-1} \] Compute the integral: \[ \int_0^1 \left( -\frac{1}{2} e^{-t^2} \right) dt = -\frac{1}{2} \int_0^1 e^{-t^2} dt = -\frac{1}{2} H(1) \] Thus, combining these results: \[ I = -\frac{1}{2} e^{-1} - \left( -\frac{1}{2} H(1) \right) = -\frac{1}{2} e^{-1} + \frac{1}{2} H(1) \] Now, substitute \( I \) back into \( H'(1) \): \[ H'(1) = -2 I = -2 \left( -\frac{1}{2} e^{-1} + \frac{1}{2} H(1) \right) = e^{-1} - H(1) \] Therefore: \[ H'(1) = e^{-1} - H(1) \] **Answer:** An explicit expression: H ′(1) = e^(–1) minus H(1); that is, H ′(1) = e^(–1) – H(1) |
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| GPT-4o | To find \( H'(1) \) for the function \( H(z) = \int_0^1 e^{-z^2 t^2} \, dt \), we need to differentiate \( H(z) \) with respect to \( z \). First, let's rewrite \( H(z) \): \[ H(z) = \int_0^1 e^{-z^2 t^2} \, dt \] To differentiate \( H(z) \) with respect to \( z \), we can use the Leibniz rule for differentiation under the integral sign. Specifically, if we have an integral of the form \( \int_{a(z)}^{b(z)} f(x, z) \, dx \), its derivative with respect to \( z \) is given by: \[ \frac{d}{dz} \left( \int_{a(z)}^{b(z)} f(x, z) \, dx \right) = \int_{a(z)}^{b(z)} \frac{\partial f(x, z)}{\partial z} \, dx + f(b(z), z) \frac{db(z)}{dz} - f(a(z), z) \frac{da(z)}{dz} \] In our case, \( a(z) = 0 \) and \( b(z) = 1 \), which are constants and their derivatives with respect to \( z \) are zero. So the derivative simplifies to: \[ H'(z) = \int_0^1 \frac{\partial}{\partial z} \left( e^{-z^2 t^2} \right) dt \] Now we need to find \( \frac{\partial}{\partial z} \left( e^{-z^2 t^2} \right) \): \[ \frac{\partial}{\partial z} \left( e^{-z^2 t^2} \right) = e^{-z^2 t^2} \cdot \frac{\partial}{\partial z} \left( -z^2 t^2 \right) = e^{-z^2 t^2} \cdot (-2z t^2) = -2z t^2 e^{-z^2 t^2} \] So, \[ H'(z) = \int_0^1 -2z t^2 e^{-z^2 t^2} \, dt \] To find \( H'(1) \), we set \( z = 1 \): \[ H'(1) = \int_0^1 -2 \cdot 1 \cdot t^2 e^{-1^2 t^2} \, dt = \int_0^1 -2t^2 e^{-t^2} \, dt \] Notice that this integral can be simplified through a substitution. Let \( u = t^2 \). Then \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). When \( t = 0 \), \( u = 0 \), and when \( t = 1 \), \( u = 1 \). Thus: \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} \, dt = \int_0^1 -2t^2 e^{-u} \cdot \frac{du}{2t} = \int_0^1 -t e^{-u} \, du = \int_0^1 -\sqrt{u} e^{-u} \, du \] However, for simplicity, we can recognize that: \[ -2 \int_0^1 t^2 e^{-t^2} \, dt = -1 \] This is by the fact that \(\int_0^\infty t^2 e^{-t^2} \, dt = \frac{\sqrt{\pi}}{4}\) and considering the limits from 0 to 1, we get the simplified result. This integral can be solved using known Gaussian integral results, but for our bounds, we can use a numerical approach or known results. Hence, we have: \[ H'(1) = -1 \] So, \( H'(1) = -1 \). |
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| llama3.1-70B-Instruct | A nice problem in calculus! To find H'(1), we need to find the derivative of H(z) with respect to z and then evaluate it at z=1. Using the Fundamental Theorem of Calculus, we can write: H(z) = ∫[0,1] e^(-z^2 t^2) dt To find the derivative of H(z), we can use the chain rule and the fact that the derivative of the exponential function is the exponential function itself: H'(z) = d/dz ∫[0,1] e^(-z^2 t^2) dt = ∫[0,1] d/dz (e^(-z^2 t^2)) dt = ∫[0,1] (-2zt^2) e^(-z^2 t^2) dt Now, we need to evaluate H'(1): H'(1) = ∫[0,1] (-2t^2) e^(-t^2) dt This integral can be evaluated using integration by parts or by recognizing that it is a Gaussian integral. Let's use the latter: H'(1) = -2 ∫[0,1] t^2 e^(-t^2) dt = -2 (1/2) ∫[0,1] e^(-t^2) dt = -∫[0,1] e^(-t^2) dt Using the error function (erf), we can write: H'(1) = -erf(1) where erf(1) ≈ 0.8427. Therefore, H'(1) ≈ -0.8427. |
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| llama3.1-405B-Instruct | To find H'(1), we need to find the derivative of H(z) with respect to z and then evaluate it at z = 1. Using the chain rule and the fact that the derivative of e^(-x) is -e^(-x), we can write: H'(z) = ∫[0,1] (-2zt)e^(-z^2t^2) dt Now, we need to evaluate H'(1). Plugging in z = 1, we get: H'(1) = ∫[0,1] (-2t)e^(-t^2) dt To evaluate this integral, we can use the substitution u = t^2, du = 2t dt. This gives: H'(1) = ∫[0,1] (-1)e^(-u) du = -∫[0,1] e^(-u) du = -[-e^(-u)] from 0 to 1 = -[-e^(-1) + 1] = e^(-1) - 1 Therefore, H'(1) = e^(-1) - 1. |
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| DeepSeek-Coder-V2 | To solve for \( H'(1) \) given the function \( H(z) = \int_0^1 e^{-z^2 t^2} dt \), we will follow these steps: 1. **Express \( H(z) \) in terms of an integral**: \[ H(z) = \int_0^1 e^{-z^2 t^2} dt \] 2. **Apply the Leibniz rule for differentiation under the integral sign**: The Leibniz rule states that if we have an integral of the form \( \int_{a(z)}^{b(z)} f(t, z) dt \), then the derivative with respect to \( z \) is: \[ \frac{d}{dz} \int_{a(z)}^{b(z)} f(t, z) dt = \int_{a(z)}^{b(z)} \frac{\partial f(t, z)}{\partial z} dt + f(b(z), z) \frac{db(z)}{dz} - f(a(z), z) \frac{da(z)}{dz} \] In our case, \( a(z) = 0 \), \( b(z) = 1 \), and \( f(t, z) = e^{-z^2 t^2} \). 3. **Compute the partial derivative of \( f(t, z) \) with respect to \( z \)**: \[ \frac{\partial f(t, z)}{\partial z} = \frac{\partial}{\partial z} e^{-z^2 t^2} = -2zt^2 e^{-z^2 t^2} \] 4. **Apply the Leibniz rule**: Since \( a(z) = 0 \) and \( b(z) = 1 \) are constants with respect to \( z \), the terms involving \( \frac{da(z)}{dz} \) and \( \frac{db(z)}{dz} \) vanish. Thus, we have: \[ H'(z) = \int_0^1 \frac{\partial}{\partial z} e^{-z^2 t^2} dt = \int_0^1 -2zt^2 e^{-z^2 t^2} dt \] 5. **Evaluate \( H'(z) \) at \( z = 1 \)**: \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} dt \] Thus, the final answer for \( H'(1) \) is: \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} dt \] |
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| Qwen2-72B-Instruct | To find the derivative of H(z) with respect to z and then evaluate it at z=1, we can use the Leibniz integral rule for differentiating under the integral sign. According to this rule, if we have an integral of the form \[ H(z) = \int_{a(z)}^{b(z)} f(t, z) dt \] the derivative of H(z) with respect to z is given by \[ H'(z) = f(b(z), z) b'(z) - f(a(z), z) a'(z) + \int_{a(z)}^{b(z)} \frac{\partial f}{\partial z}(t, z) dt \] In our case, the integral limits are constants (0 and 1), so \(a(z) = 0\), \(b(z) = 1\), and \(a'(z) = b'(z) = 0\). Therefore, the terms involving \(a'(z)\) and \(b'(z)\) vanish. The integrand is \[ f(t, z) = e^{-z^2 t^2} \] and its derivative with respect to z is \[ \frac{\partial f}{\partial z}(t, z) = -2zt^2 e^{-z^2 t^2} \] Therefore, the derivative of H(z) with respect to z is \[ H'(z) = \int_0^1 \frac{\partial f}{\partial z}(t, z) dt = \int_0^1 -2zt^2 e^{-z^2 t^2} dt \] To find \(H'(1)\), we substitute \(z = 1\) into the expression for \(H'(z)\): \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} dt \] This integral can be solved using integration by parts or recognizing it as related to the error function (erf). However, it does not have a simple closed-form solution in terms of elementary functions. Its value can be approximated numerically or expressed in terms of the error function: \[ H'(1) = -\sqrt{\pi} \frac{d}{dz} \left[ \frac{\text{erf}(zt)}{2z} \right]_{t=0}^{t=1} \Bigg|_{z=1} \] \[ H'(1) = -\sqrt{\pi} \left[ \frac{\text{erf}(1)}{2} - \frac{d}{dz} \left( \frac{\text{erf}(z)}{2z} \right) \Bigg|_{z=1} \right] \] This derivative can be further simplified using the properties of the error function, but for practical purposes, the numerical value can be found using computational tools. |
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| baichuan4 | To find $H'(z)$ we need to differentiate under the integral sign. This can be done using the Leibniz integral rule. The derivative of the integrand with respect to $z$ is $-2zte^{-z^2 t^2}$. Thus, $$H'(z) = \frac{d}{dz} \int_0^1 e^{-z^2 t^2} dt = \int_0^1 \frac{d}{dz}(e^{-z^2 t^2}) dt = \int_0^1 -2zte^{-z^2 t^2} dt.$$ Now, to find $H'(1)$, we substitute $z=1$ into the above expression: $$H'(1) = \int_0^1 -2te^{-t^2} dt.$$ This integral can be evaluated by recognizing that it is the derivative of the error function (erf), which is an odd function. Therefore, the integral from $0$ to $1$ will be half of the integral from $0$ to $\infty$. The error function is defined as $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt.$$ Taking the derivative with respect to $x$, we get $$\frac{d}{dx} \text{erf}(x) = \frac{2}{\sqrt{\pi}} e^{-x^2}.$$ Setting $x=1$, we have $$\frac{d}{dx} \text{erf}(1) = \frac{2}{\sqrt{\pi}} e^{-1}.$$ Since the integral we want is half of this value (due to the factor of $-2t$ in the integrand and the symmetry of the error function), we have $$H'(1) = -\frac{1}{\sqrt{\pi}} e^{-1}.$$ Therefore, the value of $H'(1)$ is $-\frac{1}{\sqrt{\pi}} e^{-1}$. |
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| doubao | Let $F(z,t)=e^{-z^2t^2}$. Then we have $H(z)=\int_0^1 F(z,t)dt$. Using the Leibniz rule for differentiating under the integral sign, we have: $H'(z)=\int_0^1 \frac{\partial F}{\partial z}(z,t)dt$. $\frac{\partial F}{\partial z}=-2zt^2e^{-z^2t^2}$. So $H'(1)=\int_0^1 -2t^2e^{-t^2}dt$. Evaluating this integral may require more advanced techniques or numerical methods. |
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| minimax | To find the derivative of H(z) with respect to z, we need to differentiate under the integral sign. This is possible due to Leibniz's rule for differentiating under the integral sign, which states that if we have an integral of the form: $$ H(z) = \int_{a(z)}^{b(z)} f(t, z) dt $$ then the derivative of H(z) with respect to z is: $$ H'(z) = \frac{d}{dz} \int_{a(z)}^{b(z)} f(t, z) dt = f(b(z), z) \cdot b'(z) - f(a(z), z) \cdot a'(z) + \int_{a(z)}^{b(z)} \frac{\partial}{\partial z} f(t, z) dt $$ In our case, the function H(z) is given by: $$ H(z) = \int_0^1 e^{-z^2 t^2} dt $$ Here, a(z) = 0 and b(z) = 1 are constants, so their derivatives with respect to z are zero. Therefore, the first two terms in Leibniz's rule vanish, and we are left with: $$ H'(z) = \int_0^1 \frac{\partial}{\partial z} e^{-z^2 t^2} dt $$ Now we differentiate the integrand with respect to z: $$ \frac{\partial}{\partial z} e^{-z^2 t^2} = -2z t^2 e^{-z^2 t^2} $$ So the derivative of H(z) is: $$ H'(z) = \int_0^1 -2z t^2 e^{-z^2 t^2} dt $$ Now we need to evaluate H'(1): $$ H'(1) = \int_0^1 -2 \cdot 1 \cdot t^2 e^{-1^2 t^2} dt $$ $$ H'(1) = \int_0^1 -2 t^2 e^{-t^2} dt $$ This integral does not have a simple antiderivative, so we would typically evaluate it numerically or leave it in integral form. However, if we are looking for an exact solution, we can use numerical methods or special functions to evaluate it. For the purpose of this explanation, we will leave it as an integral expression: $$ H'(1) = -2 \int_0^1 t^2 e^{-t^2} dt $$ This integral can be evaluated using numerical integration techniques such as Simpson's rule, the trapezoidal rule, or by using software that can handle numerical integration. |
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| kimi | To solve this problem, we need to find the derivative of the function H(z) with respect to z, and then evaluate it at z = 1. Given: \[ H(z) = \int_0^1 e^{-z^2 t^2} dt \] To find the derivative \( H'(z) \), we can use Leibniz's rule for differentiating under the integral sign. Leibniz's rule states that if \( F(z) = \int_{a(z)}^{b(z)} f(z, t) dt \), then \[ \frac{dF}{dz} = f(z, b(z)) \cdot b'(z) - f(z, a(z)) \cdot a'(z) + \int_{a(z)}^{b(z)} \frac{\partial f}{\partial z} dt \] In our case, \( a(z) = 0 \), \( b(z) = 1 \), and \( f(z, t) = e^{-z^2 t^2} \). Since \( a(z) \) and \( b(z) \) do not depend on z, their derivatives with respect to z are zero. Therefore, we only need to differentiate \( f(z, t) \) with respect to z. Let's find \( \frac{\partial f}{\partial z} \): \[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} e^{-z^2 t^2} \] Using the chain rule, we get: \[ \frac{\partial f}{\partial z} = e^{-z^2 t^2} \cdot (-2zt^2) \] \[ \frac{\partial f}{\partial z} = -2zt^2 e^{-z^2 t^2} \] Now, we can apply Leibniz's rule: \[ H'(z) = \int_0^1 \frac{\partial f}{\partial z} dt \] \[ H'(z) = \int_0^1 -2zt^2 e^{-z^2 t^2} dt \] Now, we need to evaluate \( H'(z) \) at \( z = 1 \): \[ H'(1) = \int_0^1 -2t^2 e^{-t^2} dt \] This integral does not have a simple closed-form solution in terms of elementary functions, but it can be evaluated using numerical methods or special functions. The integral is a form of the error function, which is defined as: \[ \text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt \] The integral we have is related to the derivative of the error function. Specifically: \[ \int_0^1 -2t^2 e^{-t^2} dt = -\frac{d}{dx} \left( \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt \right) \bigg|_{x=1} \] \[ = -\frac{d}{dx} \left( \text{erf}(x) \right) \bigg|_{x=1} \] \[ = -\text{erf}'(1) \] The derivative of the error function is given by: \[ \text{erf}'(x) = \frac{2}{\sqrt{\pi}} e^{-x^2} \] So: \[ H'(1) = -\frac{2}{\sqrt{\pi}} e^{-1} \] This is the final expression for \( H'(1) \). |
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| wenxinyiyan | 首先，我们需要计算函数$H(z)$的导数$H'(z)$。 根据题目，$H(z) = \int_0^1 e^{-z^2 t^2} dt$。 为了求$H'(z)$，我们需要对$H(z)$中的$z$求导。由于$H(z)$是一个积分形式的函数，我们需要使用积分求导法则（即莱布尼茨法则）来进行求导。 莱布尼茨法则告诉我们，如果$F(x) = \int_{a(x)}^{b(x)} f(x, t) dt$，那么$F'(x) = f(x, b(x)) \cdot b'(x) - f(x, a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) dt$。 在我们的情况下，$a(z) = 0$，$b(z) = 1$，$f(z, t) = e^{-z^2 t^2}$。因此，$a'(z) = 0$，$b'(z) = 0$。 将这些值代入莱布尼茨法则，我们得到： $H'(z) = 0 - 0 + \int_0^1 \frac{\partial}{\partial z} e^{-z^2 t^2} dt$ $= \int_0^1 \frac{\partial}{\partial z} e^{-z^2 t^2} dt$ $= \int_0^1 (-2zt^2) e^{-z^2 t^2} dt$ 现在，我们需要计算$H'(1)$。将$z = 1$代入上面的表达式，我们得到： $H'(1) = \int_0^1 (-2t^2) e^{-t^2} dt$ 这个积分可以通过使用分部积分法来求解。我们设$u = t$，$dv = -2t e^{-t^2} dt$，那么$du = dt$，$v = e^{-t^2}$。 使用分部积分法，我们得到： $H'(1) = \left[ t e^{-t^2} \right]_0^1 - \int_0^1 e^{-t^2} dt$ $= e^{-1} - \int_0^1 e^{-t^2} dt$ 最后，我们需要计算积分$\int_0^1 e^{-t^2} dt$。这个积分没有解析解，但可以通过数值方法（如梯形法则、辛普森法则等）来近似求解。 然而，在这个特定的问题中，我们只需要知道$H'(1)$的值，而不需要实际计算这个积分。因此，我们可以将$H'(1)$表示为： $H'(1) = e^{-1} - \int_0^1 e^{-t^2} dt$ 这样，我们就得到了$H'(1)$的表达式。注意，这个表达式中包含了一个无法解析求解的积分，因此在实际应用中，我们可能需要使用数值方法来近似求解这个积分。 |
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| zhipu | To find \( H'(z) \), we need to differentiate the integral \( H(z) = \int_0^1 e^{-z^2 t^2} \, dt \) with respect to \( z \). We can use the Leibniz rule for differentiation under the integral sign, which allows us to differentiate an integral with respect to a parameter that appears in the integrand. The rule states: \[ \frac{d}{dz} \left( \int_{a}^{b} f(t, z) \, dt \right) = \int_{a}^{b} \frac{\partial}{\partial z} f(t, z) \, dt \] In our case, \( f(t, z) = e^{-z^2 t^2} \). So, we need to find \( \frac{\partial}{\partial z} e^{-z^2 t^2} \): \[ \frac{\partial}{\partial z} e^{-z^2 t^2} = \frac{\partial}{\partial z} \left( e^{-z^2 t^2} \right) = -2z t^2 e^{-z^2 t^2} \] Now, applying the Leibniz rule: \[ H'(z) = \int_0^1 \frac{\partial}{\partial z} e^{-z^2 t^2} \, dt = \int_0^1 -2z t^2 e^{-z^2 t^2} \, dt \] We need to evaluate this at \( z = 1 \): \[ H'(1) = \int_0^1 -2 \cdot 1 \cdot t^2 e^{-1 \cdot t^2} \, dt = -2 \int_0^1 t^2 e^{-t^2} \, dt \] To solve this integral, we can use a substitution. Let \( u = t^2 \), then \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). When \( t = 0 \), \( u = 0 \), and when \( t = 1 \), \( u = 1 \). Substituting these into the integral: \[ -2 \int_0^1 t^2 e^{-t^2} \, dt = -2 \int_0^1 u e^{-u} \cdot \frac{du}{2\sqrt{u}} \] \[ = -2 \int_0^1 \frac{u}{2\sqrt{u}} e^{-u} \, du \] \[ = -2 \int_0^1 \frac{\sqrt{u}}{2} e^{-u} \, du \] \[ = - \int_0^1 \sqrt{u} e^{-u} \, du \] Now, let's use another substitution. Let \( v = \sqrt{u} \), then \( u = v^2 \) and \( du = 2v \, dv \). Substituting these into the integral: \[ - \int_0^1 \sqrt{u} e^{-u} \, du = - \int_0^1 v e^{-v^2} \cdot 2v \, dv \] \[ = -2 \int_0^1 v^2 e^{-v^2} \, dv \] This integral is a known form that can be evaluated using the gamma function or by recognizing it as a special integral. However, for simplicity, we can note that this integral is related to the Gaussian integral and its derivatives. The result of this integral is: \[ \int_0^1 v^2 e^{-v^2} \, dv = \frac{\sqrt{\pi}}{4} \left( 1 - \frac{1}{e} \right) \] Therefore, \[ H'(1) = -2 \cdot \frac{\sqrt{\pi}}{4} \left( 1 - \frac{1}{e} \right) \] \[ H'(1) = -\frac{\sqrt{\pi}}{2} \left( 1 - \frac{1}{e} \right) \] So, the final answer is: \[ H'(1) = -\frac{\sqrt{\pi}}{2} \left( 1 - \frac{1}{e} \right) \] |
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| gpto1 | To find \( H'(1) \), we start by differentiating \( H(z) \) with respect to \( z \): Given: \[ H(z) = \int_0^1 e^{-z^2 t^2} \, dt \] Differentiating under the integral sign (since the limits of integration are constants), we have: \[ H'(z) = \int_0^1 \frac{d}{dz} \left( e^{-z^2 t^2} \right) dt \] Compute the derivative inside the integral: \[ \frac{d}{dz} \left( e^{-z^2 t^2} \right) = e^{-z^2 t^2} \cdot (-2z t^2) = -2z t^2 e^{-z^2 t^2} \] So: \[ H'(z) = \int_0^1 \left( -2z t^2 e^{-z^2 t^2} \right) dt = -2z \int_0^1 t^2 e^{-z^2 t^2} dt \] Evaluating at \( z=1 \): \[ H'(1) = -2 \int_0^1 t^2 e^{-t^2} dt \] Now, let's compute the integral: \[ I = \int_0^1 t^2 e^{-t^2} dt \] Integrate by parts. Let: - \( u = t \), so \( du = dt \) - \( dv = t e^{-t^2} dt \), so \( v = -\frac{1}{2} e^{-t^2} \) (since \( \int t e^{-t^2} dt = -\frac{1}{2} e^{-t^2} + C \)) Using integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ I = u v \Big|_0^1 - \int_0^1 v \, du = \left( t \cdot \left( -\frac{1}{2} e^{-t^2} \right) \right) \Bigg|_0^1 - \int_0^1 \left( -\frac{1}{2} e^{-t^2} \right) dt \] Compute the boundary term: \[ \left( -\frac{1}{2} t e^{-t^2} \right) \Bigg|_0^1 = -\frac{1}{2} \cdot 1 \cdot e^{-1} - (-\frac{1}{2} \cdot 0 \cdot e^{0}) = -\frac{1}{2} e^{-1} \] Compute the integral: \[ \int_0^1 \left( -\frac{1}{2} e^{-t^2} \right) dt = -\frac{1}{2} \int_0^1 e^{-t^2} dt = -\frac{1}{2} H(1) \] Thus, combining these results: \[ I = -\frac{1}{2} e^{-1} - \left( -\frac{1}{2} H(1) \right) = -\frac{1}{2} e^{-1} + \frac{1}{2} H(1) \] Now, substitute \( I \) back into \( H'(1) \): \[ H'(1) = -2 I = -2 \left( -\frac{1}{2} e^{-1} + \frac{1}{2} H(1) \right) = e^{-1} - H(1) \] Therefore: \[ H'(1) = e^{-1} - H(1) \] **Answer:** An explicit expression: H ′(1) = e^(–1) minus H(1); that is, H ′(1) = e^(–1) – H(1) |
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