Seems Complex

You can solve a more general form of this problem and then apply it to solve this integral. Consider the following integral $\int_ {-\infty}^{\infty}{\frac{\cos{nx}}{ {({x}^{2}+{m}^{2})}^{2} }}dx$ Attempting to evaluate this directly would prove difficult. An appropriate setting for this integral is the complex plane. Let $f(z)=\frac{e^{inz}}{(z^2+m^2)^2}$ where $i$ is the imaginary unit and note that the real part of f(z) is the integrand we are interested in (this will be important later). Now, consider the contour C which is the semicircle of radius $R>b$ centered at the origin!

Let $C_R$ denote the part of C strictly above the real axis. Then $\int_C{f(z)}dz = \int_{C_R}{f(z)}dz + \int_{-R}^R{f(x)}dx$

Of these integrals, we are most interested in the rightmost one. To get its value, we will evaluate the other two and then solve for it. We will begin with the leftmost integral. In the interior of C, f(z) has one isolated singularity: namely, $z=mi$ . From the residue theorem this means that $\int_C{f(z)}dz = 2\pi i \underset{z=mi}{Res}f(z)$

To stop this solution from getting too long, I'm just going to ask that you trust that this works out to be (independent of the value of R) $\int_C{f(z)}dz = \frac{\pi(nm+1)}{2m^3e^{nm}}$

So one integral down, one to go. For the next integral, we will appeal to the Estimation Lemma. Because $\left|f(z)\right| \le \frac{1}{(R^2-n^2)^2}$ for all z in $C_R$ and the arclength of $C_R$ is $\pi R$ , we have that $\left|\int_{C_R}{f(z)}dz\right| \le \frac{\pi R}{(R^2-n^2)^2}$

If we take the limit as R approaches $\infty$ then the above integral goes towards 0. So, using the values of those two integrals we have $\frac{\pi(nm+1)}{2m^3e^{nm}} = 0 + \int_{-\infty}^{\infty}{f(x)}dx$

Earlier I made the note that the original integrand was the real part of f. This means that $\int_ {-\infty}^{\infty}{\frac{\cos{nx}}{ {({x}^{2}+{m}^{2})}^{2} }}dx = \Re \int_{-\infty}^{\infty}{f(x)}dx= \frac{\pi(nm+1)}{2m^3e^{nm}}$

In order to finally get our answer we substitute n=3 and m=5 into that equation and get $\int_ {-\infty}^{\infty}{\frac{\cos{3x}}{ {({x}^{2}+{5}^{2})}^{2} }}dx = \frac{8\pi}{125e^{15}}$

This means a=8, b=125, c=15 and a+b+c=148