# Are functions with singularities, but no poles, manipulable by Residue Calclus?

Consider these two functions

$$\int_0^{\infty} {{1} \over {(x^2+a)^{3/ 2}}} \ dx$$

And

$$\int_0^{\infty} {{1} \over {x^2+\cos(x)}} \ dx$$

They have singularities, however, wolfram alpha says that the residues at these points are 0. This seems to imply that the residue theorem alone isn't enough to evaluate the integrals.

My question is this. How would one go about using contour integration to evaluate the integral of these two functions? I'm not necessarily asking for entire derivations, just a good push in the right direction (Although feel free to evaluate them).

I think that the first integral can be evaluated by taking advantage of a branch cut along the real axis. For the second I know two zeros lie on the imaginary axis, so perhaps a quarter circle contour type could work.

• For the first integral, the integrand has branch points at $\pm \sqrt{a}$. – Mark Viola May 15 '15 at 20:29
• @Dr.MV you can create a keyhole around these points, but without a residue you'll get $0=2 \cdot \int_R f dx$. Where am I mistaken? – Zach466920 May 15 '15 at 20:32
• i think Dr. MV meant $\pm i\sqrt{a}$ – Dleep May 15 '15 at 20:34
• @EmilianoSorbello I'm aware, that doesn't really change my concern. – Zach466920 May 15 '15 at 20:38
• The second integral. Actually, the residues at the poles are not zero, merely too hard for WA to compute. – GEdgar May 15 '15 at 21:17

For the second integral, there are infinitely many poles in the upper half plane. Some $96$ of them are plotted below. These appear to be simple poles; the residue of $1/g(z)$ at a simple pole $z_0$ is $1/g'(z_0)$.
The integral should be $2\pi i$ times the sum of the residues. Of course that's an infinite series, and the terms don't have closed forms, so that might not be very useful. Numerically, $2\pi i$ times the sum of these $96$ residues is approximately $3.669413081$, while the integral itself is approximately $3.676035392$.