# integral of Gaussian function and rational function

Can any one with Mathematica help me with following integrals? O know those have closed -forms

$$\int_{-\infty}^{\infty} \frac{x^4}{(a+bx^2)^2} e^{\frac{-x^2}{2c}} dx$$

$$\int_{-\infty}^{\infty} \frac{x^3}{(a+bx^2)^2} e^{\frac{-x^2}{2c}} dx$$

a,b,c are real and positive.

• I can't say anything at first glance about the first integral, but the second integral vanishes by symmetry (integrand is an odd function). – Semiclassical Nov 16 '16 at 23:24
• The evaluation from Mathematica for the first integral is pretty awful... – Jerry Nov 16 '16 at 23:50
• One parameter among $a,b,c$ is useless. Get rid of it and switch to Fourier transforms. – Jack D'Aurizio Nov 17 '16 at 0:33

$$F(a)=\int{\frac{x^{4}}{(1+bx^{2})^{2}}e^{-ax^{2}}}dx$$

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