0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I can't say anything at first glance about the first integral, but the second integral vanishes by symmetry (integrand is an odd function). $\endgroup$ – Semiclassical Nov 16 '16 at 23:24
  • $\begingroup$ The evaluation from Mathematica for the first integral is pretty awful... $\endgroup$ – Jerry Nov 16 '16 at 23:50
  • 2
    $\begingroup$ One parameter among $a,b,c$ is useless. Get rid of it and switch to Fourier transforms. $\endgroup$ – Jack D'Aurizio Nov 17 '16 at 0:33
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.