Solution of $\int_0^{\pi} \frac{ y \cos y}{s^2+y^2} dy$

Question

Is there a solution for the following integral (even in terms of Bessel or Struve functions)? $$ \int_0^{\pi} \frac{ y \cos(y)}{s^2+y^2} \,dy $$

Answer 1

The integral can be expressed as a series.

$$ \int_0^{\pi} \frac{ y \cos(y)}{s^2+y^2} \,dy=\ln \frac{\sqrt{s^2+\pi^2}}{|s|} \cosh s+\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)!} \pi^{2n} \sum_{k=0}^{n-1} \frac{(-1)^k}{n-k} \left( \frac{s}{\pi} \right)^{2k} $$

Or, using Mathematica for the inner sum, I was able to write it using Lerch Transcendent. This series has much better convergence.

$$\int_0^{\pi} \frac{ y \cos(y)}{s^2+y^2} \,dy=\ln \frac{\sqrt{s^2+\pi^2}}{|s|} +\frac{\pi^2}{2s^2} \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)!} \pi^{2n}~ \Phi \left(-\frac{\pi^2}{s^2},1,n+1 \right)$$

See the plot for $s \in (0,10)$. Obviously, with only a few terms of the second series we can approximate the integral with very good precision:

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Just in case, how to get the first series. Expand the cosine:

$$\cos y=1-\frac{y^2}{2!}+\frac{y^4}{4!}-\cdots=\sum_{n=0}^{\infty} \frac{(-1)^n y^{2n}}{(2n)!}$$

Then we get a series of integrals in the form:

$$\int_0^{\pi} \frac{ y^{2n+1}}{s^2+y^2} \,dy=\frac{1}{2} \int_0^{\pi^2} \frac{ t^{n}}{s^2+t} \,dt$$

And finally, we can divide the fraction under the integral:

$$\frac{ t^{n}}{s^2+t}=\sum_{k=0}^{n-1} (-1)^k s^{2k} t^{n-k-1}+ \frac{(-1)^n s^{2n} }{s^2+t} $$

The last integral is elementary:

$$(-1)^n s^{2n} \int_0^{\pi^2} \frac{ dt}{s^2+t}=(-1)^n s^{2n} (\ln (s^2+\pi^2)-2 \ln s)$$