Hot Integral - 7

I solved it using the Gaussian Integral .

First find the closed form of $I_2(x)$ :

$\displaystyle I_2(x) =- \frac{8e^{x^2}}{\sqrt{\pi}} \int_0^{\infty} t^2\cos (2xt) e^{-t^2} dt$

$\displaystyle = -\frac{4e^{x^2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2\cos (2xt) e^{-t^2} dt$

$\displaystyle =- \Re \frac{4e^{x^2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2 e^{2xti} e^{-t^2} dt$

$\displaystyle =- \Re \frac{4e^{x^2}}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2 e^{2xti} e^{-t^2} dt$

$\displaystyle = -\Re \frac{4}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2 e^{x^2} e^{2xti} e^{-t^2} dt$

$\displaystyle =- \Re \frac{4}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2 e^{-(t-ix)^2} dt$

$\displaystyle \overset{{\color{grey}{t \to (t+ix) }}}{=} -\Re \frac{4}{\sqrt{\pi}} \int_{-\infty}^{\infty} (t+ix)^2 e^{-t^2} dt$

$\displaystyle = - \Re \frac{4}{\sqrt{\pi}} \bigg( {\color{#3D99F6}{\int_{-\infty}^{\infty} t^2 e^{-t^2} dt }}+ 2ix {\color{#D61F06}{\int_{-\infty}^{\infty} t e^{-t^2} dt }}+ (ix)^2 {\color{#20A900}{\int_{-\infty}^{\infty} e^{-t^2} dt}} \bigg)$

-The ${\color{#3D99F6}{\text{Blue}}}$ Integral:

Use Integration by parts with $u= t \; , \; dv = te^{-t^2} dt$ :

$\displaystyle = (t)(-\frac{1}{2}e^{-t^2} ) |_{-\infty}^{\infty} +\frac{1}{2} \int_{-\infty}^{\infty} e^{-t^2} dt$

$\displaystyle = {\color{#3D99F6}{\frac{\sqrt{\pi}}{2} }}$

-The ${\color{#D61F06}{\text{Red}}}$ Integral :

I will check its absolute convergence :

$\displaystyle \int_{-\infty}^{\infty} |t e^{-t^2} |dt = 2\int_0^{\infty} te^{-t^2} dt$

$\displaystyle \overset{{\color{grey}{t \to \sqrt{t} }}}{=} \int_0^{\infty} e^{-t} dt =1$

And hence it is absolutely convergent.

We have an odd function and opposite integration limits :

$\displaystyle ={\color{#D61F06}{0}}$

-The ${\color{#20A900}{\text{Green}}}$ Integral:

It is Simply the Gaussian Integral:

$\displaystyle = {\color{#20A900}{\sqrt{\pi}}}$

Hence:

$\displaystyle I_2(x) = - \Re \frac{4}{\sqrt{\pi}} \bigg( {\color{#3D99F6}{\frac{\sqrt{\pi}}{2} }} + 2ix({\color{#D61F06}{0}} )+ (ix)^2 {\color{#20A900}{\sqrt{\pi}}} \bigg) = 4x^2 -2$

Now solve the desired integral:

$\displaystyle k= \int_{-\infty}^{\infty} e^{-(t-1)^2} I_2(t) dt = \int_{-\infty}^{\infty} (4t^2-2) e^{-(t-1)^2} dt$

$\displaystyle \overset{{\color{grey}{t \to (t+1) }}}{=} \int_{-\infty}^{\infty} (4t^2+8t +2) e^{-t^2} dt$

$\displaystyle = 4 {\color{#3D99F6}{\int_{-\infty}^{\infty} t^2 e^{-t^2} dt }}+ 8 {\color{#D61F06}{\int_{-\infty}^{\infty} t e^{-t^2} dt }}+ 2 {\color{#20A900}{\int_{-\infty}^{\infty} e^{-t^2} dt}}$

These integrals were solved above, hence we have:

$\displaystyle k= 4({\color{#3D99F6}{\frac{\sqrt{\pi}}{2} }}) +8({\color{#D61F06}{0}}) + 2 ({\color{#20A900}{\sqrt{\pi}}} ) = 4\sqrt{\pi}$

$=> \lfloor k \rfloor = \lfloor 4\sqrt{\pi} \rfloor = \lfloor 7.08982 \rfloor = \boxed{7}$

The defining integral for $I_n$ is a standard one! We have $I_n(x) = H_n(x)$ , the $n$ th Hermite polynomial. It is easy to show that the generating function of the $I_n$ is $\sum_{n=0}^\infty I_n(x) \frac{t^n}{n!} \; =\; e^{2xt - t^2} \;,$ which is the same as the generating function of the Hermite polynomials.

The desired integral is thus $\begin{array}{rcl} \displaystyle\int_{\mathbb{R}} e^{-(t-1)^2} H_2(t)\,dt & = & \displaystyle\int_{\mathbb{R}} e^{-t^2}H_2(t+1)\,dt \; = \; \int_{\mathbb{R}} e^{-t^2} \big(4(t+1)^2 - 2\big)\,dt \\ & = &\displaystyle \int_{\mathbb{R}} e^{-t^2}\big(H_2(t) + 4H_1(t) + 4H_0(t)\big)\,dt \\ & = & \displaystyle \int_{\mathbb{R}} e^{-t^2}\big(H_2(t) + 4H_1(t) + 4H_0(t)\big)H_0(t)\,dt \\ & = & \displaystyle4\int_{\mathbb{R}} e^{-t^2} H_0(t)^2\,dt \; = \; 4\sqrt{\pi} \end{array}$ using the orthogonality properties of the Hermite polynomials.

$\displaystyle {I}_{2}(x) = - \frac{8{e}^{{x}^{2}}}{\sqrt{\pi}} \int\limits_{0}^{\infty}{{t}^{2}cos(2xt){e}^{-{t}^{2}} dt}$

$\displaystyle {I}_{2}(x) = - \frac{8{e}^{{x}^{2}}}{\sqrt{\pi}} \int\limits_{0}^{\infty}{{t}^{2}{e}^{-{t}^{2}} \sum_{k=0}^{\infty}{\frac{{(-1)}^{k}{(2xt)}^{2k}}{(2k)!}} dt}$

$\displaystyle {I}_{2}(x) = - \frac{8{e}^{{x}^{2}}}{\sqrt{\pi}} \sum_{k=0}^{\infty}{\frac{{(-1)}^{k}{(2x)}^{2k}}{(2k)!} \int\limits_{0}^{\infty}{{t}^{2+2k}{e}^{-{t}^{2}} dt}}$

Using U-substitution and Gamma function's definition, we can write it as -

$\displaystyle {I}_{2}(x) = - \frac{8{e}^{{x}^{2}}}{\sqrt{\pi}} \sum_{k=0}^{\infty}{\frac{{(-1)}^{k}{(2x)}^{2k}}{(2k)!}\Gamma\left(k+1+\frac{1}{2}\right)}$

Using Gamma function property,

$\displaystyle = - \frac{8{e}^{{x}^{2}}}{\sqrt{\pi}} \sum_{k=0}^{\infty}{\frac{{(-1)}^{k}{(2x)}^{2k}}{(2k)!}\frac{(2k+2)!\sqrt{\pi}}{{2}^{2k+2}(k+1)!}}$

Simplifying,

$\displaystyle = - 2{e}^{{x}^{2}} \sum_{k=0}^{\infty}{\frac{{(-{x}^{2})}^{k}(2k+1)}{k!}}$

Using the Taylor series of ${e}^{x}$ and then differentiation, one can get

$\displaystyle = -2{e}^{{x}^{2}}\left(-{e}^{-{x}^{2}}(2x^2-1)\right)$

$\displaystyle = 4{x}^{2}-2$

Then, the concluding integral becomes

$\displaystyle \int_{-\infty}^{\infty}{(4{t}^{2}-2){e}^{-{(1-t)}^{2}} dt}$

whose computation will be the same as that of Hasan.