Complex integration exercise 2

Question

I have to calculate $$\int^\infty_{-\infty} \frac {\cos x}{x^2+x+\frac 5 4}dx.$$ If I use the function $\frac {\cos z}{z^2+z+\frac 5 4} $ over the contour $A=[-R,R]\cup Re^ {i\theta} $, with $\theta \in [0,\pi] $, I can't show that the integral over $Re^ {i\theta}$ goes to zero as $R $ increases; so I don't know how to proceed. If I use the function Re$(e^{iz})$ always over $A $, I don't know how to calculate precisely the residue in the pole (in $i-\frac 1 2$); clearly that value is equal to the integral requested. Is there another way to proceed, or should I insist with one of the integration above? Thanks in advance

Answer 1

Using the $e^{iz}$ approach, you can compute the residue at $z=-1/2+i$ thus,

$$f(z)=\frac{e^{iz}}{z^2+z+\frac{5}{4}}=\frac{e^{iz}}{(z-(-1/2+i))(z-(-1/2-i))}.$$ Using partial fractions, $$f(z)=\left(\frac{i/2}{z-(-1/2-i)}-\frac{i/2}{z-(-1/2+i)}\right)e^{iz},$$ hence $$\text{Res}(f,-1/2+i)=-\frac{i}{2}e^{i(-1/2+i)}.$$

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The reason you may choose to use $e^{iz}$ instead of $\cos z$ is because we get the convergence we need, or at least convergence to something which may contribute to the final evaluation of the original integral. For example, in your case $$\oint_\Gamma\frac{e^{iz}}{z^2+z+\frac{5}{4}}dz=2\pi i\sum\text{residues}=2\pi i\left(-\frac{i}{2}e^{i(-1/2+i)}\right),$$ where $\Gamma=[-R,R]\cup\gamma$ where $\gamma=\{Re^{i\theta}\mid \theta\in[0,\pi]\}$. We can split the integral up into piecewise paths, i.e. $$\int_{-R}^R\frac{e^{iz}}{z^2+z+\frac{5}{4}}dz+\int_\gamma \frac{e^{iz}}{z^2+z+\frac{5}{4}}dz=\pi e^{i(-1/2+i)}.$$ Now let $R\to\infty$, then we end up with $$\int_{-\infty}^\infty \frac{e^{iz}}{z^2+z+\frac{5}{4}}dz+\lim_{R\to\infty}\int_\gamma \frac{e^{iz}}{z^2+z+\frac{5}{4}}dz=\pi e^{i(-1/2+i)}.$$ Now, since $e^{iz}=\cos(z)+i\sin(z)$, and we know that real and imaginary parts are always independent of each other, then we can choose to take the real parts of the equality only. This produces the original integral we were studying in the left hand side, which we can isolate algebraically to get to an answer! Note that in general the integral over $\gamma$ may not always be zero.

In this case, it turns out the integral over $\gamma$ does vanish ! So we obtain $$\int_{-\infty}^\infty \frac{\cos z}{z^2+z+\frac{5}{4}}dz=\Re\int_{-\infty}^\infty \frac{e^{iz}}{z^2+z+\frac{5}{4}}dz=\Re\pi e^{i(-1/2+i)}=\frac{\pi \cos \left(\frac{1}{2}\right)}{e}.$$