On the Laplace Transform of $\frac{1}{1+\sin^2(x)}$

Question

Are there any methods of evaluating this integral?$$I(s)=\int_0^\infty \frac{e^{-sx}}{1+\sin^2(x)}dx$$ I tried expanding the denominator into a geometric series to get $$I(s)=\sum_{n=0}^\infty (-1)^n\int_0^\infty \sin^{2n}(x)e^{-sx}dx$$ However I am not sure of how to go about evaluating this integral. Is this the right approach or are there any better ways of finding a closed form for $I(s)$?

Answer 1

EDIT

I have added a second paragraph with some results for the approach of expanding the denominator.

Original post

Let

$$I(s) = \int_0^\infty \frac{e^{-s x}}{1+\sin(x)^2}\,dx$$

§1. As a first step we derive a formula for $I(s)$ with a definite integral.

To this end we decompose the integration region in equidistant intervals of size $\pi$.

$$I(s) = \sum_{k=0}^\infty \int_{k \pi}^{(k+1)\pi} \frac{e^{-s x}}{1+\sin(x)^2}\,dx$$

Making use of the periodicity of the denominator of the integrand we can write, using $x\to k \pi + y$

$$\int_{k \pi}^{(k+1)\pi} \frac{e^{-s x}}{1+\sin(x)^2}\,dx \\ = e^{- k s \pi}\int_{0}^{\pi} \frac{e^{-s y}}{1+\sin(y)^2}\,dy $$

Now the geometric sum can be done giving

$$I(s) = \frac{1}{1-e^{-\pi s}}g(s)$$

with

$$g(s) = \int_{0}^{\pi} \frac{e^{-s y}}{1+\sin(y)^2}\,dy$$

$2. Expanding the denominator (added 4.4.19)

As proposed in the OP we expand the denominator in a geometric series.

The integrals to be computed are

$$a_k (s) =(-1)^k \int_0^\infty e^{-s x}\sin(x)^{2k}\,dx$$

The result can be written as

$$a_k(s) = \frac{1}{s} \frac{(-1)^k (2 k)!}{\prod _{m=1}^k \left((2 m)^2+s^2\right)}$$

Notice that $a_k(s)$ has simple poles at $s=\pm 2 m i, m=0..k$.

Expressed via the Pochhammer symbol we have

$$a_k(s)=\frac{\left(-\frac{1}{4}\right)^k (2 k)!}{s \left(1-\frac{i s}{2}\right)_k \left(\frac{i s}{2}+1\right)_k}$$

The asymptotic behaviour for large $k$ is

$$a_k(s) \simeq \frac{1}{{\sqrt{\pi } s}} (-1)^k \sqrt{\frac{1}{k}} \Gamma \left(1-\frac{i s}{2}\right) \Gamma \left(\frac{i s}{2}+1\right)\\=\frac{1}{2} \sqrt{\pi } (-1)^k \sqrt{\frac{1}{k}} \text{csch}\left(\frac{\pi s}{2}\right) $$

This shows that the (alternating) series is convergent.

An approximation for $I(s)$ can be found by taking the first term $a_0(s) = \frac{1}{s}$ and adding the sum of this asymtotic expression, starting a $k=1$:

$$I(s) \simeq \frac{1}{s}+\frac{1}{2} \left(\sqrt{2}-1\right) \sqrt{\pi } \zeta \left(\frac{1}{2}\right) \text{csch}\left(\frac{\pi s}{2}\right)$$

A graphical comparision of $I(s)$ with the approximation is shown here