Calculation of integral $\int\exp \left(-\alpha \sin^2 \left(\frac{x}{2} \right) \right) dx$

Question

Given $\alpha$ is a constant. How to calculate the following integral?

\begin{equation} \int \exp \bigg(-\alpha \sin^2 \bigg(\frac{x}{2} \bigg) \bigg) dx \end{equation}

Thanks for your answer.

Answer 1

The indefinite integral cannot be expressed in terms of elementary functions. See Liouville's theorem and the Risch algorithm for more details. However, for $n\in$ Z we have the following identity in terms involving Bessel functions:

$$\int_0^{n\pi}e^{-\alpha\sin^2\Big(\tfrac x2\Big)}dx=\frac{n\cdot\pi}{\sqrt{e^\alpha}}\cdot\text{Bessel I}_0\bigg(\frac\alpha2\bigg)$$

Answer 2

A series solution is given via the Jacobi-Anger expansion: $$-\alpha\, \left( \sin \left( \frac{1}{2}\,x \right) \right) ^{2}=\frac{1}{2}\,\alpha \, \left( \cos \left( x \right) -1 \right) $$ $${{\rm e}^{\frac{1}{2}\,\alpha\, \left( \cos \left( x \right) -1 \right) }}={ {\rm e}^{-\frac{1}{2}\,\alpha}}\,{{\rm I}_0\left(\frac{1}{2}\,\alpha\right)}+2\,{ {\rm e}^{-\frac{1}{2}\,\alpha}}\sum _{n=1}^{\infty } {{\rm I}_n\left(\frac{1}{2}\,\alpha\right)}\cos \left( nx \right)$$ where: $${i}^{n}{{\rm J}_n\left(-\frac{1}{2}\,i\alpha\right)}= {{\rm I}_n\left(\frac{1}{2}\,\alpha\right)}$$ was used to relate the Bessel function of the first kind $J$ to the modified Bessel function $I$. Integration term by term gives: $$\int \!{{\rm e}^{\frac{1}{2}\,\alpha\, \left( \cos \left( x \right) -1 \right) }}{dx}={{\rm e}^{-\frac{1}{2}\,\alpha}} {{\rm I}_0\left(\frac{1}{2}\,\alpha\right)}x+2\,{{\rm e}^{-\frac{1}{2}\,\alpha}} \sum _{n=1}^{\infty }{\frac {{{\rm I}_n\left(\frac{1}{2}\alpha\right)} \sin \left( nx \right) }{n}}$$

For integration between zero and integer multiples of $\pi$ the sum clearly vanishes and you recover @Lucian's solution.

Answer 3

Let $u=\dfrac{x}{2}$ ,

Then $x=2u$

$dx=2~du$

$\therefore\int e^{-\alpha\sin^2\frac{x}{2}}~dx$

$=2\int e^{-\alpha\sin^2u}~du$

$=2\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}du$

$=2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$

For $n$ is any natural number,

$\int\sin^{2n}u~du=\dfrac{(2n)!u}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

$\therefore2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$

$=2u+2\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n(2n)!u}{4^n(n!)^3}-2\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^3(2k-1)!}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!2u}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!x}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}\dfrac{x}{2}\cos\dfrac{x}{2}}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$

Or you can express in terms of the incomplete bessel K function:

$\int e^{-\alpha\sin^2\frac{x}{2}}~dx$

$=\int e^\frac{\alpha(\cos x-1)}{2}~dx$

$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha\cos x}{2}~dx$

$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha(e^{ix}+e^{-ix})}{4}~dx$

$=e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$

Let $u=e^{-ix}$ ,

Then $x=i\ln u$

$dx=\dfrac{i}{u}du$

$\therefore e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$

$=ie^{-\frac{\alpha}{2}}\int\dfrac{e^{\frac{\alpha u}{4}+\frac{\alpha}{4u}}}{u}du$

$=ie^{-\frac{\alpha}{2}}\int_0^u\dfrac{e^{\frac{\alpha t}{4}+\frac{\alpha}{4t}}}{t}dt+C$

$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{ut}d(ut)+C$

$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{t}dt+C$

$=ie^{-\frac{\alpha}{2}}\int_\infty^1te^{\frac{\alpha u}{4t}+\frac{\alpha t}{4u}}~d\left(\dfrac{1}{t}\right)+C$

$=ie^{-\frac{\alpha}{2}}\int_1^\infty\dfrac{e^{\frac{\alpha t}{4u}+\frac{\alpha u}{4t}}}{t}dt+C$

$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha}{4u},-\dfrac{\alpha u}{4}\right)+C$ (according to http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html)

$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha e^{ix}}{4},-\dfrac{\alpha e^{-ix}}{4}\right)+C$