Integration FTW!

I won't be giving justification of my steps,

Firstly I am proving some results here :

$\displaystyle \int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx } = \sqrt { \pi }$

$\displaystyle \int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 } }dx } =\frac { \sqrt { \pi } }{ 2 }$

Proof :

With a little change of variables first integral got converted into :

$\displaystyle \int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 } }dx } =2\int _{ 0 }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 } }dx } =\int _{ 0 }^{ \infty }{ \sqrt { t } { e }^{ -t }dt } =\Gamma (3/2)=\frac { 1 }{ 2 } \Gamma (1/2)=\frac { \sqrt { \pi } }{ 2 }$

Similarly second can be proved.

Now back to our integral we have

$\displaystyle I = \frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ { -x }^{ 2 } }(1+cos(2\pi x))dx }$

We will be computing $\displaystyle J = \int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 } }cos(2\pi x)dx } =Re(\int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 }+2\pi ix }dx } )$

$\Rightarrow J = Re({ e }^{ -{ \pi }^{ 2 } }\int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ (x-\pi i) }^{ 2 } }dx } )$

$t= x-\pi i$ we have ;

$\displaystyle J = Re({ e }^{ -{ \pi }^{ 2 } }\int _{ -\infty }^{ \infty }{ { (t+\pi i) }^{ 2 }{ e }^{ -{ t }^{ 2 } }dt } )$

$\displaystyle J = Re({ e }^{ -{ \pi }^{ 2 } }\int _{ -\infty }^{ \infty }{ { t }^{ 2 }{ e }^{ -{ t }^{ 2 } }+2\pi it{ e }^{ -{ t }^{ 2 } }-{ \pi }^{ 2 }{ e }^{ -{ t }^{ 2 } }dt } )$

Now observe that $\displaystyle \int _{ -\infty }^{ \infty }{ 2\pi ix{ e }^{ -{ x }^{ 2 } }dx } =0$

Hence our integral becomes :

$\displaystyle J = { e }^{ -{ \pi }^{ 2 } }\int _{ -\infty }^{ \infty }{ { x }^{ 2 }{ e }^{ -{ x }^{ 2 } }-{ \pi }^{ 2 }{ e }^{ -{ x }^{ 2 } }dx }$

Using our results proved above we have :

$J = \dfrac { \sqrt { \pi } { e }^{ { -\pi }^{ 2 } } }{ 2 } (1-2{ \pi }^{ 2 })$

Putting the value of $J$ in $I$ we have finally the value of $I$ as :

$I = \dfrac { \sqrt { \pi } { e }^{ { -\pi }^{ 2 } } }{ 4 } (1-2{ \pi }^{ 2 })+\dfrac { \sqrt { \pi } }{ 4 }$

I am going to use a general result $\int_{-\infty}^{\infty} e^{-x^{2}} dx = \sqrt{\pi}$ . Let $\Large I(a) =\int_{-\infty}^{\infty} e^{-ax^{2}}cos^{2}(x\pi) dx$ So $\Large 2I(a) =\int_{-\infty}^{\infty} e^{-ax^{2}}(1+cos(2x\pi)) dx$ Let $I_{1} = \int_{-\infty}^{\infty}e^{-ax^{2}} dx$ and $I_{2} = \int_{-\infty}^{\infty}e^{-ax^{2}}cos(2x\pi) dx$ Let us evaluate $I_{1}$ first:- To do that use the substitution $\sqrt{a}x = t$ To get $I_{1} = \sqrt{\frac{\pi}{a}}$

$I_{2}$ can be written as :-

$\Large I_{2} = \Re \int_{-\infty}^{\infty}e^{-ax^{2}}e^{-2ix\pi} dx$

$\Large I_{2} = \Re( e^{-\frac{\pi^{2}}{a}}\int_{-\infty}^{\infty}e^{-(ax^{2}+2ix\pi - \frac{\pi^{2}}{a})} dx)$ $\Large I_{2} = \Re( e^{-\frac{\pi^{2}}{a}}\int_{-\infty}^{\infty}e^{-(\sqrt{a}x + \frac{i\pi}{\sqrt{a}})^{2}} dx)$

Substitute $\sqrt{a}x + \frac{i\pi}{\sqrt{a}} = t$ . To get

$\Large I_{2} = e^{-\frac{\pi^{2}}{a}} \sqrt{\frac{\pi}{a}}$

$\Large 2I(a) = \sqrt{\frac{\pi}{a}}(1+e^{-\frac{\pi^{2}}{a}})$

Now as we can see if we apply Leibnitz Rule for differnetiation under the integral sign. We see that $\Large I'(a) =-\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}}cos^{2}(x\pi) dx$ So we need $-I'(1)$ Differentiating the expression we got for $I(a)$ we have:-

$\Large 2I'(a) = -\frac{\sqrt{\pi}}{2a^{\frac{3}{2}}}(1+e^{-\frac{\pi^{2}}{a}}) + \sqrt\frac{\pi}{a}e^{-\frac{\pi^2}{a}}\frac{\pi^{2}}{a^{2}}$

Substitute $a=1$ to get the answer as $0.4426$

Disambiguation:- $\Re(.)$ denotes the real part of a complex number.

An alternate solution. First, write $\cos^2(\pi x)$ as a series.

$\begin{aligned} \cos^2(\pi x) &= \frac{1}{2}\left[\cos(2\pi x) + 1\right]\\ &= \frac{1}{2}\left[\left(\sum_{n=0}^{\infty} (-1)^n\frac{(2\pi x)^{2n}}{(2n)!}\right)+ 1\right]\\ \end{aligned}$

Next plug that into the integral, and exchange the sum and integral signs (this is valid because $x^{2n}e^{-x^2}$ is nonnegative everywhere; monotone convergence theorem).

$\begin{aligned} \int_{-\infty}^{\infty}x^2e^{-x^2}\cos^2(\pi x)\,dx&=\int_{-\infty}^{\infty}x^2e^{-x^2}\frac{1}{2}\left[\left(\sum_{n=0}^{\infty} (-1)^n\frac{(2\pi x)^{2n}}{(2n)!}\right)+ 1\right]\,dx\\ &=\frac{1}{2}\left[\left(\sum_{n=0}^{\infty} (-1)^n\frac{(2\pi)^{2n}}{(2n)!}\right)\int_{-\infty}^{\infty} x^{2(n+1)}e^{-x^2} dx \right]+\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{-x^2} dx\\ &=\frac{1}{2}\left[\left(\sum_{n=0}^{\infty} (-1)^n\frac{(2\pi)^{2n}}{(2n)!}\right) \left(n+\frac{1}{2}\right)!\right]+\frac{\sqrt{\pi}}{4}\\ &=\frac{\sqrt{\pi}}{4}\left[1+\sum_{n=0}^{\infty} \frac{(-1)^n(2\pi)^{2n}}{2^{2n}}\frac{\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)}{n\left(n-\frac{1}{2}\right)(n-1)\cdots \left(\frac{3}{2}\right)(1)\left(\frac{1}{2}\right)}\right]\\ &=\frac{\sqrt{\pi}}{4}\left[1+2\sum_{n=0}^{\infty}\left(-\pi^2\right)^n\frac{\left(n+\frac{1}{2}\right)}{n!}\right]\\ &=\frac{\sqrt{\pi}}{4}\left[1+2(1-\pi^2)\sum_{n=0}^{\infty}\frac{\left(-\pi^2\right)^n}{n!}\right]\\ &=\frac{\sqrt{\pi}}{4}\left[1+2\left((\frac{1}{2}-\pi^2\right)e^{-\pi^2}\right]\\ &= 0.4427\ldots \end{aligned}$

I am not giving full solution here. Just giving sketch of my solution : Take, $f(a) = \int_{\infty}^{\infty} x^2e^{-x^2}\cos^2(ax)$ 1.) Compute $f'(a)$

2.) Apply parts

3.) Make differential equation

4.) Solve it

5.) Use $f(0) = \frac{\sqrt{\pi}}{2}$

6.) Find $f(\pi)$

P.S: IF you didn't got anything, then ignore. I will post complete solution when i get time. [ After may -_- ]