How can I evaluate
$$\int_{-a}^a \frac{\cos(\pi n x/a)}{x^2+a^2}\,\mathrm dx$$?
My attempt :
$$\begin{align}\mathcal{I}& =\displaystyle\int_{-a}^a \dfrac{\cos(\pi n x/a)}{x^2+a^2}\mathrm dx \\ & = 2\displaystyle\int_0^a \dfrac{\cos(\pi n x/a)}{x^2+a^2}\, \text{ ,even integrand}\\ &=2\displaystyle\int_0^{\pi n} \dfrac{\cos u}{\frac{a^2u^2}{n^2\pi^2}+1}\left(\dfrac{a\mathrm du}{n\pi}\right)\, \text{ ,via substituting $u=\dfrac{\pi nx}{a}$}\\ &=\dfrac{2\pi n}{a}\displaystyle\int_0^{\pi n} \dfrac{\cos u}{u^2+n^2\pi^2}\, \mathrm du\\ & =\dfrac{2\pi n}{a}\displaystyle\int_0^{n\pi} \dfrac{\cos u}{(u-i\pi n)(u+i\pi n)}\,\mathrm du\\ & =\dfrac{i}{a}\underbrace{\displaystyle\int_0^{n\pi} \dfrac{\cos u}{u+i\pi n}\mathrm du}_{\mathcal{I_1}}-\dfrac{i}{a}\overbrace{\displaystyle\int_0^{n\pi}\dfrac{\cos u}{u-in\pi}\mathrm du}^{\mathcal{I_2}}\, \text{ ,via partial fraction decomposition}\end{align}$$
$$\begin{align}\mathcal{I_1}&=\displaystyle\int_0^{n\pi}\dfrac{\cos u}{u+i\pi n}\mathrm du \\&=\displaystyle\int_0^{(i+1)n\pi}\dfrac{\cos(v-i\pi n)}{v}\mathrm dv\, \text{ ,via substituting $v=u+i\pi n$}\\ & = \displaystyle\sum_{k=0}^\infty \dfrac{(-1)^k}{(2k)!}\displaystyle\int_0^{(i+1)n\pi} \dfrac{(v-i\pi n)^{2k}}{v}\mathrm dv\, \text{ ,via Maclaurin series}\end{align}$$
from where I can't figure out further.
Any help is appreciated.
Best Answer
For $$I=\int\dfrac{\cos( u)}{u+i\pi n}\,du $$ $$u+i\pi n=v \implies I=\int \frac{\cos(v-i\pi n)}v \,dv$$ Expand the cosine $$\cos(v-i\pi n)=\cosh (\pi n) \cos (v)+i \sinh (\pi n) \sin (v)$$ $$I=\cosh (\pi n) \int \frac{\cos(v)}v \,dv+i \sinh (\pi n) \int \frac{\sin(v)}v \,dv$$ $$I=\cosh (\pi n)\,\text{Ci}(v)+i \sinh (\pi n) \,\text{Si}(v)$$