# Math Help - differentiating in the integral sign

1. ## 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.

2. Originally Posted by galactus
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

3. I always like your use of the gamma, Mathstud. I like it too.

4. $\frac{1}{x^a}\int_0^{\infty}z^{a-1}e^{-z}dz=\frac{\Gamma(a)}{x^{a}}$.

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.

5. Originally Posted by galactus
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 $\frac{1}{x^a}=\frac{1}{\Gamma(a)}\int_0^{\infty}z^ {a-1}e^{-xz}dz$