The integral $\int_{-\infty}^{\infty}\frac{1}{R^2+z^2}e^{-\alpha \sqrt{R^2+z^2}}\mathrm{d}z$ I need to evaluate the following integral (for a physical application):
$$I=\int_{-\infty}^{\infty}\frac{1}{R^2+z^2}e^{-\alpha \sqrt{R^2+z^2}}\mathrm{d}z, $$
where $\alpha>0$ and $R>0$.
I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $\alpha$, which does not make sense physically.
 A: 
In this answer, we use the following well-known general results:
$$K_{n}(z)=\frac{\sqrt{\pi}}{\Gamma(n+\frac{1}{2})}(z/2)^n\int_1^{\infty} \frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx \tag{1}$$
And:
$$\int K_0(z)~dz=\frac{\pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C \tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.

We use differentiation under the integral sign:
$$I(\alpha)=\int_{-\infty}^{\infty} \frac{1}{R^2+z^2}e^{-\alpha\sqrt{R^2+z^2}}~dz$$
$$I'(\alpha)=-\int_{-\infty}^{\infty} \frac{e^{-\alpha\sqrt{R^2+z^2}}}{\sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=\sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$\begin{align}\int_{-\infty}^{\infty} \frac{e^{-\alpha\sqrt{R^2+z^2}}}{\sqrt{R^2+z^2}}~dz&=2\int_{0}^{\infty} \frac{e^{-\alpha\sqrt{R^2+z^2}}}{\sqrt{R^2+z^2}}~dz\\&=2\int_R^{\infty} \frac{e^{-at}}{\sqrt{t^2-R^2}}~dt\\&=2\int_1^{\infty} \frac{e^{-\alpha Ru}}{\sqrt{u^2-1}}~du  \end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(\alpha)=-2K_0(\alpha R)$$
We can now integrate with respect to $\alpha$, using equation $(2)$. We obtain:
$$I(\alpha)=-\pi\alpha(L_{-1}(\alpha R)K_0(\alpha R)+L_0(\alpha R)K_1(\alpha R))+C$$
Using that $I(0)=\int_{-\infty}^{\infty} \frac{1}{R^2+z^2}~dz=\pi/R$, we obtain that $C=\pi/R$. Thus, the result is:
$$I=\frac{\pi}{R}-\pi\alpha(L_{-1}(\alpha R)K_0(\alpha R)+L_0(\alpha R)K_1(\alpha R))$$
I highly doubt that one can simplify this further.
A: *

*Differentiate w.r.t $\alpha$

*Sub $z=R \sinh(x)$

*do the integral over x $\int_{\mathbb{R}}dx\exp(-c\cosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

*integrate back w.r.t. $\alpha$
Use  $\tfrac{2}{\pi z}\int dz  K_0(z)= L_0(z)
    K_0(z)+L_{-1}(z)K_1(z)+C$, $L_b(z)$ is a so called Struve
function

*Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $\alpha=\infty$
PS: Your contour Approach fails since you didn't take the branchcut $[i,i\infty]$ into account :/
A: When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.
