Integral $\int_0^{2\pi}e^{\sin^2x}dx$

Question

My attempt at a solution using the substitution method: $$u=\sin^2x\\du=2\sin x\cos xdx\\du=2u^{\frac12}{(1-\sin^2x)}^{\frac12}dx\\du=2u^{\frac12}{(1-u)}^{\frac12}dx\\\int_0^{2\pi}e^{\sin^2x}dx\\\int_0^1e^{u}\frac{1}{2u^{\frac12}{(1-u)}^{\frac12}}du$$ I have no idea how to solve it. thanks in advance.

Answer 1

1. Solution with contour integraton

Do the following substitution

$$ \sin x = \frac{z-z^{-1}}{2i}$$

$$ dx = \frac{dz}{iz}$$

Hence

$$I= \int_{0}^{2\pi} e^{\sin^2 x} dx = \frac{1}{i} \oint_{|z|=1} \frac{e^{\left(\frac{z^2-1}{2iz}\right)^2}}{z} dz$$

Your integral become a contour integral around the unit complex circle in the poistive sense.

The function:

$$ f(z) = \frac{e^{\left(\frac{z^2-1}{2iz}\right)^2}}{z}$$

Has a singularity at the origin $z=0$. Hence:

$$I = \frac{1}{i} \oint_{|z|=1} \frac{e^{\left(\frac{z^2-1}{2iz}\right)^2}}{z} dz = 2\pi \operatorname{Res}(f,0)$$

To find the residue expand the function in the Laurent series:

$$ e^{\left(\frac{z^2-1}{2z}\right)^2} = \sum_{n=0}^{\infty} \left(\frac{z^2-1}{2iz}\right)^{2n} \frac{1}{n!}$$

$$ \left(z^2-1\right)^{2n} = \sum_{j=0}^{2n} \binom{2n}{j} (-1)^{2n-j} z^{2j} $$

$$e^{\left(\frac{z^2-1}{2iz}\right)^2} = \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \sum_{j=0}^{2n} \binom{2n}{j} \frac{(-1)^{n-j}z^{2j-2n}}{n!}$$

$$f(z) = \frac{e^{\left(\frac{z^2-1}{2z}\right)^2}}{z} = \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \sum_{j=0}^{2n} \binom{2n}{j} \frac{(-1)^{n-j}z^{2j-2n-1}}{n!}$$

The residue is the coefficient of $ z^{2j-2n-1} = z^{-1} \Longrightarrow j=n $

$$\therefore \operatorname{ Res}(f,0) = \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \binom{2n}{n}\frac{1}{n!}$$

$$\boxed{I = 2\pi \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \binom{2n}{n}\frac{1}{n!}}$$

This sum converges to $\displaystyle 2\pi \sqrt{e}I_{0}\left(\frac{1}{2}\right)$

Update 1:

I found a way to evaluate this last series:

Note $$ \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \binom{2n}{n}\frac{1}{n!} = \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \frac{(2n)!}{n!^3 }$$

Now recall that

$$ (2n)! = (1)_{2n}$$

where $(x)_{n}=x(x+1)(x+2)\cdots(x+n-1)$ is the rising factorial or Pochhammer polynomial

Now using the duplication formula: $\displaystyle (x)_{2n} = 4^n\left(\frac{x}{2}\right)_{n}\left(\frac{x+1}{2}\right)_{n}$ and the fact that $(1)_{n} = n!$

We have

$$ (2n)! = (1)_{2n} = 4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}$$

Hence $$ \sum_{n=0}^{\infty} \frac{1}{2^{2n}} \binom{2n}{n}\frac{1}{n!} = \sum_{n=0}^{\infty} \frac{4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}}{2^{2n} (1)_{n}(1)_{n} n!} = \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{2}\right)_{n}}{(1)_{n}}\frac{1}{n!} ={}_1F_{1}\left(\frac{1}{2};1;1\right)$$

Using the identity 3 of Generalization of Kummer's transformations here

$$e^{-\frac x 2} \, {}_1F_1(a,2a,x)= {}_0F_1 \left (;a+\tfrac 1 2; \tfrac{x^2}{16} \right ) \tag{*}$$

with $\displaystyle x= 1$ and $\displaystyle a= \frac{1}{2}$

$${}_1F_{1}\left(\frac{1}{2};1;1\right) = \sqrt{e} \;{}_0F_{1}(;1,\frac{1}{16})$$

But $\displaystyle \sqrt{e}\; {}_0F_{1}(;1,\frac{1}{16})$ is the hypergometric representation of $\displaystyle I_{0}\left(\frac{1}{2}\right)$: $$ {}_{0}F_{1}(;1,\frac{1}{16}) = \sum_{n=0}^{\infty} \frac{1}{4^{2n}n!^2} = I_{0}\left(\frac{1}{2}\right)$$

Therefore

$$\boxed{ I =\int_{0}^{2\pi} e^{\sin^2 x} dx= 2\pi \sqrt{e} \; I_{0}\left(\frac{1}{2}\right) }$$

2. Solution using the complete beta function(Update 2)

Let $n\in \mathbb{N}$. Recall:

$$\int_{0}^{2\pi} \sin^{2n}\theta d\theta= 4\int_{0}^{\frac{\pi}{2}} \sin^{2n}\theta d\theta \stackrel{w=\cos\theta}{=} 4\int_{0}^{1} w^{2n}(1-w^2)^{-\frac{1}{2}}dw \stackrel{r=w^2}{=}2\int_{0}^{1} r^{n+\frac{1}{2}-1}(1-r)^{\frac{1}{2}-1}dr = 2B\left(n+\frac{1}{2},\frac{1}{2}\right) = \frac{2\Gamma\left(n+\frac{1}{2}\right)\sqrt{\pi}}{\Gamma(n+1)} = \frac{2\pi}{2^{2n}}\binom{2n}{n}$$

Therefore:

$$\int_{0}^{2\pi} \sin^{2n}\theta d\theta =\frac{2\pi}{2^{2n}}\binom{2n}{n}$$

Summing both sides of the equation:

$$\sum_{n=0}^{\infty}\frac{1}{n!}\int_{0}^{2\pi} \sin^{2n}\theta d\theta = 2\pi \sum_{n=0}^{\infty}\frac{1}{2^{2n}n!}\binom{2n}{n}$$

Applying the Fubini-Tonelli theorem in the left hand side we can interchange series and integral sign:

$$\int_{0}^{2\pi} \sum_{n=0}^{\infty}\frac{\sin^{2n} \theta}{n!} d\theta = 2\pi \sum_{n=0}^{\infty}\frac{1}{2^{2n}n!}\binom{2n}{n}$$

$$\Longrightarrow \int_{0}^{2\pi} e^{\sin^2\theta}d\theta = 2\pi \sum_{n=0}^{\infty}\frac{1}{2^{2n}n!}\binom{2n}{n}$$

Applying the identity (*) from above

$$\boxed{ I =\int_{0}^{2\pi} e^{\sin^2 x} dx= 2\pi \sqrt{e} \; I_{0}\left(\frac{1}{2}\right) }$$

Answer 2

According to Wolfram, it is

$$\int_{0}^{2\pi} e^{[\sin x]^2}\ dx = 2\sqrt{e} \pi I_0\left(\frac{1}{2}\right)$$

I think there isn't really a good "trick" for this beyond just recognizing that the modified Bessel functions have well-known integral identities involving integrands of a broadly similar shape in that they involve $e^{\cos x}$, so one should then try to mash them to see if one can get something to come out. Namely,

$$I_\alpha(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos \alpha \theta\ d\theta - \frac{\sin \alpha \pi}{\pi} \int_{0}^{\infty} e^{-x \cosh t - \alpha t}\ dt$$

is a "known quantity" in that it's published already, and note that if one takes $\alpha = 0$, the right-hand term dies and the left becomes

$$I_0(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} d\theta$$

hence use the half-angle formula $[\sin x]^2 = \frac{1 - \cos 2x}{2}$ so

$$\int_{0}^{2\pi} e^{[\sin x]^2}\ dx = \int_{0}^{2\pi} e^{\frac{1 - \cos 2x}{2}}\ dx = \int_{0}^{2 \pi} e^{1/2} e^{-\frac{1}{2} \cos{2x}}\ dx = \sqrt{e} \int_{0}^{2\pi} e^{-\frac{1}{2} \cos 2x}\ dx$$

then you should be able to take it from there with scalings.