Integration of exponential and square root function I need to solve this $$\int_{-\pi}^{\pi} \frac{e^{ixn}}{\sqrt{x^2+a^2}}\,dx,$$ 
where $i^2=-1$ and $a$ is a constant.
 A: This is not complete yet.
$$\int_{-\pi}^{\pi} \dfrac {\exp(inx)}{\sqrt{x^2+a^2}}dx=\int_{-\pi}^{\pi}\dfrac {\cos(nx)}{\sqrt{x^2+a^2}}dx+i \int_{-\pi}^{\pi}\dfrac {\sin(nx)}{\sqrt{x^2+a^2}}dx=2\int_{0}^{\pi}\dfrac {\cos(nx)}{\sqrt{x^2+a^2}}dx=2I$$
Set $x=a \sinh y$ then $dx=(a \cosh y) dy$ and $\sqrt{x^2+a^2}=a \cosh y$. Let $\pi=a \sinh y_0$.
Then the integration $I$ becomes:
$$I =\int_{0}^{y_0}\cos(n a \sinh y)dy$$
A: By definition, $\displaystyle\int_0^\infty\frac{\cos x}{\sqrt{x^2+a^2}}dx=K_0\big(|a|\big)$, where K is a Bessel function. Letting $x=nt$, we 
have $\displaystyle\int_0^\infty\frac{\cos(nt)}{\sqrt{t^2+a^2}}dt=K_0\big(|an|\big)$. Unfortunately, there are no “incomplete” Bessel functions, 
so your integral does not possess a closed from even in terms of such special expressions, unless, of 
course, $n=0$, in which case the answer is simply $\text{arcsinh}\dfrac\pi{|a|}$ . As an aside, for positive values of a 
and n, we have $\displaystyle\int_0^\infty\frac{\sin(nx)}{\sqrt{x^2+a^2}}dx=\frac\pi2\Big[I_0(an)-L_0(an)\Big]$, where I is another Bessel function, 
and L is a Struve function.
