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Math Help - differentiating in the integral sign

  1. #1
    Eater of Worlds
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    differentiating in the integral sign

    I am sure most of you are familiar with what is known as Feynman integration.

    For those who are not, it allows us to integrate otherwise difficult integrals. Ones we would use

    contour integration on. We choose a parameter. A 'suitable dominating

    function' which we can use to make the integrand friendlier.

    Example:

    Let's integrate \int_{0}^{\frac{\pi}{2}}xcot(x)dx

    With this one it is not immediately seen where we can use our parameter.

    But, if we rewrite it as follows:

    I(a)=\int_{0}^{\frac{\pi}{2}}\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}dx, we can do this:

    I'(a)=\frac{d}{dx}\int_{0}^{\frac{\pi}{2}}\frac{ta  n^{-1}(a\cdot tan(x))}{tan(x)}dx=\int_{0}^{\frac{\pi}{2}}\frac{{  \partial}}{{\partial}b}\left[\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}\right]dx

    =\int_{0}^{\frac{\pi}{2}}\frac{dx}{(a\cdot tan(x))^{2}+1}

    =\frac{\pi}{2(a+1)}

    Integrate w.r.t a gives us:

    I(a)=\frac{\pi}{2}ln(a+1)

    Now, leaving a=1, we get:

    \int_{0}^{\frac{\pi}{2}}xcot(x)dx=I(1)=\frac{\pi}{  2}ln(2)

    Which, by the way, gives the result of another famous one:

    \int_{0}^{\frac{\pi}{2}}ln(sin(x))dx=-\int_{0}^{\frac{\pi}{2}}xcot(x)dx=\frac{-\pi}{2}ln(2)

    I know, this is Leibniz. But it is a cool way to integrate. the trick is finding the parameter.

    It works pretty slick on:

    \int_{0}^{\pi}e^{cos(x)}cos(sin(x))dx

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    How about trying it on the famous Fresnel integrals:

    \int_{0}^{\infty}cos(x^{2})dx

    or

    \int_{0}^{\infty}sin(x^{2})dx

    by choosing the proper parameter.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    I am sure most of you are familiar with what is known as Feynman integration.

    For those who are not, it allows us to integrate otherwise difficult integrals. Ones we would use

    contour integration on. We choose a parameter. A 'suitable dominating

    function' which we can use to make the integrand friendlier.

    Example:

    Let's integrate \int_{0}^{\frac{\pi}{2}}xcot(x)dx

    With this one it is not immediately seen where we can use our parameter.

    But, if we rewrite it as follows:

    I(a)=\int_{0}^{\frac{\pi}{2}}\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}dx, we can do this:

    I'(a)=\frac{d}{dx}\int_{0}^{\frac{\pi}{2}}\frac{ta  n^{-1}(a\cdot tan(x))}{tan(x)}dx=\int_{0}^{\frac{\pi}{2}}\frac{{  \partial}}{{\partial}b}\left[\frac{tan^{-1}(a\cdot tan(x))}{tan(x)}\right]dx

    =\int_{0}^{\frac{\pi}{2}}\frac{dx}{(a\cdot tan(x))^{2}+1}

    =\frac{\pi}{2(a+1)}

    Integrate w.r.t a gives us:

    I(a)=\frac{\pi}{2}ln(a+1)

    Now, leaving a=1, we get:

    \int_{0}^{\frac{\pi}{2}}xcot(x)dx=I(1)=\frac{\pi}{  2}ln(2)

    Which, by the way, gives the result of another famous one:

    \int_{0}^{\frac{\pi}{2}}ln(sin(x))dx=-\int_{0}^{\frac{\pi}{2}}xcot(x)dx=\frac{-\pi}{2}ln(2)

    I know, this is Leibniz. But it is a cool way to integrate. the trick is finding the parameter.

    It works pretty slick on:

    \int_{0}^{\pi}e^{cos(x)}cos(sin(x))dx

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    How about trying it on the famous Fresnel integrals:

    \int_{0}^{\infty}cos(x^{2})dx

    or

    \int_{0}^{\infty}sin(x^{2})dx

    by choosing the proper parameter.
    \int\sin\left(x^2\right)dx Make the sub x^2=\varphi to give \frac{1}{2}\int_0^{\infty}\frac{\sin\left(\varphi\  right)}{\sqrt{\varphi}}

    Now consider hte gamma function \Gamma(a)=\int_0^{\infty}t^{a-1}e^{-t}dt. Now consider the integral \int_0^{\infty}t^{a-1}e^{-xy}dx. Letting xt=z this itnegral becomes \int_0^{\infty}\frac{1}{x^{a-1}}z^{a-1}e^{-z}\frac{dz}{x}=\frac{1}{x^a}\int_0^{\infty}z^{a-1}e^{-z}dz=\frac{\Gamma(a)}{x^{a}}.

    Now how is this useful?

    \frac{\Gamma(a)}{x^a}=\int_0^{\infty}z^{a-1}e^{-xz}dz\implies\frac{1}{x^a}=\frac{1}{\Gamma(a)}\int  _0^{\infty}z^{a-1}e^{-xz}dz

    So we can use this parameter to rewrite our integral as

    \frac{1}{2\Gamma\left(\frac{1}{2}\right)}\int_0^{\  infty}\int_0^{\infty}\frac{e^{-xz}\sin(x)}{\sqrt{z}}dxdz

    From there it is simple
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  3. #3
    Eater of Worlds
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    I always like your use of the gamma, Mathstud. I like it too.
    Last edited by galactus; December 23rd 2008 at 12:28 PM.
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  4. #4
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    \frac{1}{x^a}\int_0^{\infty}z^{a-1}e^{-z}dz=\frac{\Gamma(a)}{x^{a}}.

    Now how is this useful?

    \frac{\Gamma(a)}{x^a}=\int_0^{\infty}z^{a-1}\overbrace{e^{-xz}}^{\text{here}}dz\implies\frac{1}{x^a}=\frac{1}  {\Gamma(a)}\int_0^{\infty}z^{a-1}e^{-xz}dz

    So we can use this parameter to rewrite our integral as

    \frac{1}{2\Gamma\left(\frac{1}{2}\right)}\int_0^{\  infty}\int_0^{\infty}\frac{e^{-xz}\sin(x)}{\sqrt{z}}dxdz

    From there it is simple
    I got to looking at this in depth and have a question or two. How did you just add that x in the e^{-xz}. Is

    there some identity I am overlooking?. How did you actually use it to rewrite the integral?. I am not seeing something.

    If \frac{1}{x^{a}}\int_{0}^{\infty}z^{a-1}e^{-z}dz=\frac{{\Gamma}(a)}{x^{a}}, then how can:

    \frac{{\Gamma}(a)}{x^{a}}=\int_{0}^{\infty}z^{a-1}e^{-xz}dz?. Just wondering.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    I got to looking at this in depth and have a question or two. How did you just add that x in the e^{-xz}. Is

    there some identity I am overlooking?. How did you actually use it to rewrite the integral?. I am not seeing something.

    If \frac{1}{x^{a}}\int_{0}^{\infty}z^{a-1}e^{-z}dz=\frac{{\Gamma}(a)}{x^{a}}, then how can:

    \frac{{\Gamma}(a)}{x^{a}}=\int_{0}^{\infty}z^{a-1}e^{-xz}dz?. Just wondering.
    I showed it I thought?

    \int_0^{\infty}z^{a-1}e^{-xz}. Let xz=\varphi this becomes \int_0^{\infty}\left(\frac{\varphi}{x}\right)^{a-1}e^{-\varphi}\frac{d\varphi}{x}=\frac{1}{x^a}\int_0^{\i  nfty}\varphi^{a-1}e^{-\varphi}=\frac{\Gamma(a)}{x^a}

    So \frac{\Gamma(a)}{x^a}=\int_0^{\infty}z^{a-1}e^{-xz}dz then dividing both sides by \Gamma(a) gives \frac{1}{x^a}=\frac{1}{\Gamma(a)}\int_0^{\infty}z^  {a-1}e^{-xz}dz
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