How to evaluate $\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$ How to evaluate the following integral
$$\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x$$
Can we solve it without using complex analysis method?
 A: I would use the identity 
$$(1+X)^{-\gamma}=\frac{1}{\Gamma(\gamma)}\int_0^\infty d\xi\ \xi^{\gamma-1}e^{-\xi}e^{-\xi X}$$
to write
$$
\int_{0}^{\infty }\frac{x^{4}e^{-2x^{2}}}{\left ( 1+x^{2} \right )^{4}}\mathrm{d}x
=\frac{1}{\Gamma(4)}\int_{0}^\infty d\xi\ \xi^3 e^{-\xi} \int_0^\infty dx\ x^4 e^{-2 x^2-\xi x^2}=\frac{1}{\Gamma(4)}\int_0^\infty d\xi \xi^3 e^{-\xi}\frac{3 \sqrt{\pi }}{8 (\xi +2)^{5/2}}\ ,
$$
which seems much nicer. Change variables to $\xi+2=t$ to get
$$
\frac{3\sqrt{\pi}e^{2}}{8\Gamma(4)}\int_2^\infty dt (t-2)^3 \frac{e^{-t}}{t^{5/2}}\ .
$$
Expanding $(t-2)^3$, you are left with four elementary integrals. 
A: Define
$$
f_k(a)=\int_0^\infty\frac{e^{-ax^2}\,\mathrm{d}x}{\left(1+x^2\right)^k}\tag{1}
$$
Then we have
$$
\int_0^\infty\frac{x^4e^{-2x^2}}{(1+x^2)^4}\,\mathrm{d}x=f_4''(2)\tag{2}
$$
From $(1)$, we get the recursion $f_k(a)-f_k'(a)=f_{k-1}(a)$, which means
$$
f_k(a)=e^a\int_a^\infty e^{-t}f_{k-1}(t)\,\mathrm{d}t\tag{3}
$$
Starting with
$$
f_0(a)=\frac12\sqrt{\frac\pi{a}}\tag{4}
$$
and applying $(3)$ $4$ times, we get
$$
f_4(a)=\frac{\sqrt{\pi a}}{48}\left(4a^2-8a+15\right)-\frac{\pi e^a}{96}\left(8a^3-12a^2+18a-15\right)\operatorname{erfc}\left(\sqrt{a}\right)\tag{5}
$$
Applying $(2)$ to $(5)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{x^4e^{-2x^2}}{(1+x^2)^4}\,\mathrm{d}x
=\frac{17}{16}\sqrt{2\pi}-\frac{241}{96}\pi e^2\operatorname{erfc}\left(\sqrt2\right)}\tag{6}
$$
where
$$
\operatorname{erfc}(a)=\frac2{\sqrt\pi}\int_a^\infty e^{-x^2}\,\mathrm{d}x\tag{7}
$$
A: Lets apply Plancherel's theorem 
$$
\int_{\mathbb{R}} f(x)g(x)dx=
\int_{\mathbb{R}}\hat{f}(k)\hat{g}(k)dk \quad (*)
$$
Here the hat stands for the  Fourier transform of the corresponding quantity
Note that this is possible because we are dealing with an even function and we can drop some compelx conjugations because everything is real.
Let's set $f(x)=e^{-2x^2}$ and $g(x)=\frac{x^4}{(1+x^2)^4}$.
Then we have (both FT are quiet standard,complete the square for the first and use residue theorem for the second)
$$
\hat{f}(k)=\frac{1}{2}e^{-k^2/8}\\
\hat{g}(k)=\frac{\sqrt{\pi}}{48\sqrt{2}}\left(\Theta(k)e^{-k}(k^3-6 k^2+3k+3)-\Theta(-k)e^{k}(k^3+6 k^2+3k-3)\right)
$$
where $\Theta(k)$ denotes the Heaviside function. I don't have the time a the moment to finish the rather tedious but straightforward calculation of (*) from here on but i'm sure u can bring it home if u keep in mind the definition of the error function
A: We will illustrate a methodology for evaluating the integral of interest that is based on "Feynmann's Method" for differentiating under the integral.  
To begin, we augment the integral of interest by introducing the parameter $a$ in the exponential argument of the integrand.  let $I(a)$ denote the integral 
$$I(a)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx\tag 1$$
Then, the integral of interest is simply given by
$$I(2)=\int_0^\infty \frac{x^4e^{-ax^2}}{(1+x^2)^4} \,dx$$
Using partial fraction expansion on the term $\frac{x^4e^{-ax^2}}{(1+x^2)^4}$ reveals
$$\frac{x^4}{(1+x^2)^4}=\frac{1}{(1+x^2)^2}-\frac{2}{(1+x^2)^3}+\frac{1}{(1+x^2)^4} \tag 2$$
whereby we can write $(1)$ as
$$\begin{align}
I(a)&=\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^2}\,dx-2\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^3}\,dx+\int_0^\infty \frac{e^{-ax^2}}{(1+x^2)^4}\,dx \\\\
&=e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx-2e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^3}\,dx+e^a\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^4}\,dx \tag 3
\end{align}$$
We can evaluate each of the $3$ integrals on the right-hand side of $(3)$ by differentiating under the integral.  In particular, we will evaluate the first integral on the right-hand side of $(3)$ and leave the evaluation of the second and third integrals as an exercise.
Let $F_n(a)$ denote the integral
$$F_n(a)=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^n}\,dx$$
The integral $F_0(a)$ is Gaussian and easily evaluated; we find that $F_0(a)$ is 
$$\begin{align}
F_0(a)&=\int_0^\infty e^{-a(1+x^2)}\,dx\\\\
&=\frac{\sqrt \pi}{2}\frac{e^{-a}}{\sqrt a}\end{align}$$

INTEGRATING $F_0(a)$
Noting $F_1'(a)=-F_0(a)$, with $F_1(0)=\pi/2$, yields
$$\begin{align}
F_1(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{1+x^2}\,dx\\\\
&=\frac\pi 2-\int_0^a F_0(y)\,dy\\\\
&=\frac{\pi }{2}-\frac{\sqrt \pi}{2}\int_0^a \frac{e^{-y}}{\sqrt y}\,dy\\\\
&=\frac{\pi}{ 2}-\frac{\pi}{2}\frac{2}{\sqrt \pi}\int_0^a e^{-t^2}\,dt\\\\
&=\frac{\pi}{2}\text{erfc}(\sqrt a)
\end{align}$$

INTEGRATING $F_1(a)$
Similarly, since $F_2'(a)=-F_1(a)$, with $F_2(0)=\pi/4$, we have
$$\begin{align}
F_2(a)&=\int_0^\infty \frac{e^{-a(1+x^2)}}{(1+x^2)^2}\,dx\\\\
&=\frac\pi 4-\int_0^a F_1(x)\,dx\\\\
&=\frac\pi 4-\frac{\pi}{2}\int_0^a \text{erfc}(\sqrt x)\,dx\\\\
&=\frac\pi 4-\pi \int_0^{\sqrt a} x\,\text{erfc}(x)\,dx\\\\
&=\frac\pi 4-\pi\left(\frac a2\text{erfc}(\sqrt a)-\frac12 \int_0^{\sqrt{a}} x^2\,\text{erfc}'(x)\,dx\right)\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\int_0^\sqrt a x^2e^{-x^2}\,dx\\\\
&=\frac\pi 4-\frac\pi 2 a\,\text{erfc}(\sqrt a)-\frac\pi 2 \frac{2}{\sqrt \pi}\left(\frac{\sqrt \pi}{4}\text{erf}(\sqrt a)-\frac {\sqrt a}2  e^{-a}\right)\\\\
&=\frac{\pi}{4}\left(1-2a\right)\text{efrc}(\sqrt a)+\frac{\sqrt \pi}{2}\sqrt a e^{-a}
\end{align}$$
Thus, the first term on the right-hand side of $(3)$ is $e^2F_2(2)$.
The evaluation of $F_3(2)$ and $F_4(2)$ follow along the same methodology and is left as an exercise.
