What is the value of $\int_{-\infty}^{\infty} \frac{e^{-ix}}{x^{2 }+ 4} dx$

• Calculate the Fourier transform of $b(x) =\frac{1}{x^2 +a^2}$ [closed]

Solutions Collecting From Web of "What is the value of $\int_{-\infty}^{\infty} \frac{e^{-ix}}{x^{2 }+ 4} dx$"

Let $f$ be the meromorphic function on $\mathbb C$ below ($a>0$):

$$z\mapsto \frac{e^{-iaz}}{z^2+4}$$

It has two simple poles, both of order 1. They are the roots of $z^2+4$, i.e. $2i$ and $-2i$.

The idea is to use the residue theorem, but of course you need to pick the contour carefully. Due the nature of the denominator of $f$, we clearly want to pick some contour with a semicircle, since our integral is calculated on $\mathbb R$, we also want a segment on the real axis.

Two such simple contours seem possible, i.e. $$\gamma_R^+=\sigma_R+\theta^+_R$$

or $$\gamma_R^-=\sigma_R+\theta_R^-$$

with $$\sigma_R\colon[-R,R]\ni t\mapsto t$$
$$\theta_R^\pm:[0,\pi]\ni t\mapsto \pm Re^{it}$$

In this case, the better choice is $\gamma_R^-$ as we will see below.

If $R>2$, then the only pole in the interior of $\gamma_R^-$ is $-2i$.
The winding number of $\gamma_R^-$ about this pole is $-1$, since we’re turning clockwise around it.

Hence, by the residue theorem, we get for $R>2$ :

$$\oint_{\gamma_R^-}f(z)\ \mathrm{d}z = -2i\pi\ \mathrm{Res}(f,-2i)$$

The residue is easy to calculate :
$$\mathrm{Res}(f,-2i)=\lim_{z\to -2i} (z+2i)f(z)=\frac{e^{-2a}}{-4i}$$

So $$\oint_{\gamma_R^-} f(z)\ \mathrm{d}z=\frac\pi2 e^{-2a}$$

We don’t really see for the moment why $\gamma_R^-$ is a better choice than $\gamma_R^+$.

We’ll see why when $R$ will tend to $+\infty$.

$$\oint_{\gamma_R^-} f(z)\ \mathrm{d}z=\int_{\sigma_R} f(z)\ \mathrm{d}z+\int_{\theta_R^-} f(z)\ \mathrm{d}z$$

The left term of the RHS will tend to the integral we want to evaluate because $x\mapsto \dfrac{e^{-iax}}{x^2+4}$ is integrable.

$$\lim_{R\to +\infty} \int_{\sigma_R} f(z)\ \mathrm{d}z=\lim_{R\to +\infty} \int_{-R}^R \frac{e^{-iax}}{x^2+4}\ \mathrm{d}x=\int_{-\infty}^{+\infty} \frac{e^{-iax}}{x^2+4}\ \mathrm{d}x$$

Now to the right term. If $R>2$, then :
$$\left|\int_{\theta_R^-} f(z)\ \mathrm{d}z\right|=\left|\int_0^\pi -iRe^{it}\frac{e^{iaRe^{it}}}{R^2e^{2it}+4}\ \mathrm{d}t\right|\leqslant \frac{R}{R^2-4} \int_0^\pi e^{-Ra\sin(t)}\ \mathrm{d}t$$

Now the choice of $\gamma_R^-$ is evident. With $\gamma_R^+$, we would have find almost the same results, with only one minor change : the integral at the rightmost member above would have been $\displaystyle\int_0^\pi e^{Ra\sin(t)}\ \mathrm{d}t$ instead, and this is not clear that this integral would tend to $0$ with the $\frac{R}{R^2-4}$ close to it since the integrand can take arbitrary large value ($e^R$ for example).

In our case, we can notice that :

$$\int_0^\pi e^{-Ra\sin(t)}\mathrm{d} t=2\int_0^{\frac\pi2} e^{-Ra\sin(t)}\ \mathrm{d}t$$

Using the concavity of $\sin$ on $\left[0,\frac\pi2\right]$, we get :

$$\forall t\in\left[0,\frac\pi2\right], \sin(t)\geqslant \frac{2t}\pi$$

Hence

$$\int_0^{\frac\pi2} e^{-Ra\sin(t)}\ \mathrm{d}t\leqslant \int_0^{\frac\pi2} e^{-\frac{2Rat}\pi}\ \mathrm{d}t=\frac\pi{2Ra}(1-e^{-2Ra})\to 0\quad (R\to +\infty)$$

Since we also have
$$\lim_{R\to +\infty} \frac{R}{R^2-4} = 0$$

we find that :

$$\lim_{R\to +\infty} \int_{\gamma_R^-} f(z)\ \mathrm{d}z = 0$$

Finally, we get :

$$\int_{-\infty}^{+\infty} \frac{e^{-aix}}{x^2+4}\ \mathrm{d} x = \frac \pi 2 e^{-2a}\quad (a>0)$$

For the general case, also check the great answers given here. (Thanks to Arthur for this link).