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Thread: How take the derivative of this integral?

  1. #1
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    Question How take the derivative of this integral?

    $$T(x,t) = \frac{1}{(4 \pi \alpha t)^{1/2}} \int_{x'=0}^{\infty} F(x')\Bigg\lbrace\exp \left [ -\frac{(x-x')^2}{4 \alpha t}\right ] + \exp \left [ -\frac{(x+x')^2}{4 \alpha t}\right ] \Bigg\rbrace dx'$$


    how can we calculate


    $$\frac{dT}{dt}\Bigg |_{x=0}$$
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  2. #2
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    Re: How take the derivative of this integral?

    T is a function of the two variables, x and t, so I presume you mean the partial derivative of T with respect to t. Start by using the product rule. We have $\displaystyle \frac{1}{4\pi\alpha}t^{-1/2}$ times $\displaystyle \int_{x'= 0}^\infty F'(x')\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$.

    I presume you know that the derivative of $\displaystyle t^{-1/2}$ is $\displaystyle -\frac{1}{2}t^{-3/2}$ so this derivative is $\displaystyle -\frac{1}{8\pi\alpha}t^{-3/2}\int_{x'= 0}^\infty F'(x')\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$ plus $\displaystyle \frac{1}{4\pi\alpha}t^{-1/2}$ times the derivative of that integral with respect to t.

    Since t does not appear in either limit of integration, the derivative of the integral is just $\displaystyle \int_{x'= 0}^\infty F'(x')\frac{\partial }{\partial t}\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$.
    Thanks from kimia
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  3. #3
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    Re: How take the derivative of this integral?

    Quote Originally Posted by HallsofIvy View Post
    T is a function of the two variables, x and t, so I presume you mean the partial derivative of T with respect to t. Start by using the product rule. We have $\displaystyle \frac{1}{4\pi\alpha}t^{-1/2}$ times $\displaystyle \int_{x'= 0}^\infty F'(x')\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$.

    I presume you know that the derivative of $\displaystyle t^{-1/2}$ is $\displaystyle -\frac{1}{2}t^{-3/2}$ so this derivative is $\displaystyle -\frac{1}{8\pi\alpha}t^{-3/2}\int_{x'= 0}^\infty F'(x')\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$ plus $\displaystyle \frac{1}{4\pi\alpha}t^{-1/2}$ times the derivative of that integral with respect to t.

    Since t does not appear in either limit of integration, the derivative of the integral is just $\displaystyle \int_{x'= 0}^\infty F'(x')\frac{\partial }{\partial t}\left(e^{-\frac{(x-x')^2}{4\alpha}t^{-1}+ e^{-\frac{(x+x')^2}{4\alpha}t^{-1}\right)dx'$.
    Thanks for your reply.

    1. Since we are interested in $x=0$, I assumed that the PDE is transformed to an ODE. Am I wrong?

    2. Your math code does not display for me. Could you please check it out as I can fully follow your approach. Thanks again.
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