Computation of a certain contour integral

Question

I have to do the following integral(using complex analysis):

$$\int_{0}^{\infty} \frac{\cos{nx}}{x^{4}+1} dx $$

So, first I evaluated $x^{4}+1=0 $ and got $x = \pm \frac{1+i}{\sqrt{2}}, \pm \frac{-1+i}{\sqrt{2}} $

Let, $\Gamma$ be a semicircle(in the upper half-plane) of radius $R$, where $R > 1$ so,

$$ \int_{\Gamma} \frac{\cos{nz}}{z^{4}+1} dz = \int_{\Gamma} \frac{cos(nz)}{(z-z_{1})(z-z_{2})(z-z_{3})(z-z_{4})} dz$$, where $z_{1}, z_{2}, z_{3}, z_{4}$ are the four zeroes of $ z^{4} +1 = 0$. Then I computed the residues in the following manner:

Res($f$, $\frac{1+i}{\sqrt{2}})$ = $ \lim_{\ z\to\ \frac{1+i}{\sqrt{2}}} \frac{cosnz}{(z+\frac{1+i}{\sqrt{2}})((z-\frac{-1+i}{\sqrt{2}} )(z+\frac{-1+i}{\sqrt{2}})} $

Since, the expression is messy I will call the residue $a$ and doing the same thing to compute the other residue I get $b$.

Then, we notice that on the semicircular arc:

$$ \left|\frac{cos(nz)}{z^{4} +1} \right| \leq \frac{1}{R^{4}+1} $$ and

$$ \left| \int \frac{cos(nz)}{z^{4} +1} dz \right| \leq \frac{1}{R^{4}+1} .\pi R $$

So, the integral goes to $0$ as $R \rightarrow \infty$. So, as $R \rightarrow \infty$

$$\int_{\Gamma} f(z) dz = \int_{-\infty}^{\infty} \frac{\cos{nx}}{x^{4}+1} dx = 2\pi i(a+b)$$

So,

$$ \int_{0}^{\infty} \frac{\cos{nx}}{x^{4}+1} dx = \frac{1}{2} \int_{0}^{\infty} \frac{\cos{nx}}{x^{4}+1} dx = \frac{1}{2}(a+b) \pi i $$

Is my work correct? If not, where did I go wrong?

Answer 1

Some things:

• You want to apply Cauchy's integral theorem, so your contour must be closed. You have a semicircle, but you do not mention the linear segment closing it into the boundary of a half-disk. You need this segment to be part of your contour, as it is the part that actually is (twice) the integral you want.
• Why do you believe $|\cos(n z)|$ is bounded by $1$ on the semicircular part of $\Gamma$ (as the radius increases)? $\cos(n z) = \frac{1}{2}\left( \mathrm{e}^{\mathrm{i} n z} + \mathrm{e}^{-\mathrm{i} n z} \right)$, which increases like $\mathrm{e}^{|z|}$ along the positive imaginary axis. (Sure, the real cosine is bounded by $1$, but this is the complex cosine.)
  
  The exponential eventually overpowers any polynomial in the denominator.
  
• Because of the rapid growth away from the real axis, you don't want your contour to go very far from the real axis. Note that $\cos(x + \mathrm{i} y) = \cos x \cosh y - \mathrm{i} \sin x \sinh y$. You might have more luck with two parallel rays, one above the pole in the first quadrant, one along the positive x-axis, with a short line segment descending the imaginary axis to join the ends of the rays. (Just a guess : I haven't checked that this is "easy".)
• Also, you are integrating a real integrand over a real interval. If you don't get a real result, something has gone terribly wrong.