Can any one with Mathematica help me with following integrals? O know those have closed -forms
\begin{equation} \int_{-\infty}^{\infty} \frac{x^4}{(a+bx^2)^2} e^{\frac{-x^2}{2c}} dx \end{equation}
\begin{equation} \int_{-\infty}^{\infty} \frac{x^3}{(a+bx^2)^2} e^{\frac{-x^2}{2c}} dx \end{equation}
a,b,c are real and positive.
$$F(a)=\int{\frac{x^{4}}{(1+bx^{2})^{2}}e^{-ax^{2}}}dx$$
note that $F(a\rightarrow0)\rightarrow\infty;F(a\rightarrow\infty)\rightarrow0$
rewrite as
$$F(a)=\int{\frac{x^{4}}{(1-b\frac{d}{da})^{2}}e^{-ax^{2}}}dx$$
and by some abuse
$$F(a)=\frac{1}{(1-b\frac{d}{da})^{2}}\int{x^{4}e^{-ax^{2}}}dx=\frac{1}{(1-b\frac{d}{da})^{2}}\frac{3}{4}\sqrt{\pi}a^{-5/2}$$
further abuse
$$(1-b\frac{d}{da})^{2}F(a)=\frac{3}{4}\sqrt{\pi}a^{-5/2}$$
To solve
$$F=u(a)e^{a/b}$$
$$(1-b\frac{d}{da})^{2}F(a)=b^{2}u''(a)e^{a/b}=\frac{3}{4}\sqrt{\pi}a^{-5/2}$$
hence $u''=\sqrt{\pi}\frac{3}{4b^{2}}e^{-a/b}a^{-5/2}$
lets ignore the constants and consider
$u''=e^{-a/b}a^{-5/2}$
$$u=aK_{1}+K_{2}+\int_{a}^{\infty}da'\int_{a'}^{\infty}da''e^{-a^{''}/b}a^{''-5/2}$$
due to the $a\rightarrow\infty$ limit we deduce that $K_{1};K_{2}$ are zero
$$\int_{a}^{\infty}da'\int_{a'}^{\infty}da''e^{-a^{''}/b}a^{''-5/2}=\int_{a}^{\infty}da^{''}\int_{a}^{a^{''}}da^{'}e^{-a^{''}/b}a^{''-5/2}=\int_{a}^{\infty}da^{''}\left[a^{''}-a\right]e^{-a^{''}/b}a^{''-5/2}$$
$$=\int_{a}^{\infty}da^{''}e^{-a^{''}/b}a^{''-3/2}-a\int_{a}^{\infty}da^{''}e^{-a^{''}/b}a^{''-5/2}$$
$$=b^{-1/2}\Gamma(-1/2,a/b)-ab^{-3/2}\Gamma(-3/2,a/b)$$
so finally
$$F(a)=u(a)e^{a/b}=\sqrt{\pi}\frac{3}{4b^{3/2}}e^{a/b}\left[\Gamma(-1/2,a/b)-ab^{-1}\Gamma(-3/2,a/b)\right]$$
hopefully I didn't miss anything
