0

I need to solve the following integral: \begin{align*} \int_{0}^{\pi} \sin^2(\theta) e^{-\frac{c}{\sin^4 \theta}} d\theta \end{align*} I tried a few changes of variables to make it look like a Gaussian integral, as well as integration by parts, but could not get it to a manageable form, mainly because the $sin^4$ term is in the denominator of the exponent. I would appreciate any ideas.

karakusc
  • 1,512
  • There are a lot of integrals without closed form, what makes you think that it has a closed form? – Elaqqad Dec 06 '16 at 21:03
  • Nothing in particular. If an exact closed form is not feasible, a close upper bound would be useful too. – karakusc Dec 06 '16 at 21:13

1 Answers1

3

Use symmetry to write this as twice the integral from $0$ to $\pi/2$, then substitute $t = \sin(\theta)^{-1/4}$ to make it $$ \int_1^\infty \frac{e^{-ct^2}\; dt}{t^2 \sqrt{t-1}}$$ But I don't think there's a closed form for this, unless you allow hypergeometric functions: Maple gives $$ {\frac {24\,{c}^{5/4}{\mbox{$_2$F$_2$}(\frac34,\frac54;\,\frac32,\frac94;\,-c)} \left( \Gamma \left( 3/4 \right) \right) ^{2}-20\,\sqrt {2}\pi\, {\mbox{$_2$F$_2$}(\frac14,\frac34;\,\frac12,\frac74;\,-c)}{c}^{3/4}+15\,\pi\,\Gamma \left( 3/4 \right) }{30 \;\Gamma \left( 3/4 \right) }} $$

This corresponds to the series $$4\,{c}^{5/4}\sum _{k=0}^{\infty }{\frac {\Gamma \left( 3/4+ k \right) }{ \left( 5+4\,k \right)(1+2k)! } \left(-4c \right) ^{k}}-2\,c^{3/4}\,\sum _{k =0}^{\infty }{\frac {\Gamma \left( 1/4+k \right) \left( -4\,c \right) ^{k}}{ \left( 3+4\,k \right)(2k)! }}+\frac{\pi}{2} $$

so for $c$ not too big, partial sums of this will give good approximations.

On the other hand, for large $c$ you can get good approximations using Watson's lemma.

Robert Israel
  • 448,999