Stuck on this integration $\int_0 ^\infty \frac{1}{1+x^2} \cos(kx) dx =\frac{\pi}{2}e^{-k}$ I'm not sure how to show this
$$\int_0 ^\infty \frac{1}{1+x^2} \cos(kx) \ \mathrm dx =\frac{\pi}{2}e^{-k}$$
I tried by parts but I'm not getting anywhere, I'd really appreciate the help
 A: (N.B.: throughout, I assume $k>0$; since the integral is clearly an even function of $k$ for real $k$, this is sufficient.)
I would say you have four main options here:


*

*Use the Fourier inversion theorem: we know that the Fourier transform of a function has a unique inverse. This carries over to the cosine transform as
$$ \int_0^{\infty} f(x) \cos{kx} \, dx = F(k) \iff \int_0^{\infty} F(k) \cos{kx} \, dk = \pi f(x) $$
The unique continuous function on the positive real axis with Fourier transform $1/(1+x^2)$ is $e^{-k}/2$, and the inversion theorem above gives the result. (You may call this a bit cheap, of course... I was lucky in knowing what transformed to $1/(1+x^2)$.)

*Use differentiation under the integral sign: (this way is also going to be mildly illegal) Write
$$ I(k) = \int_0^{\infty} \frac{\cos{kx}}{1+x^2} \, dx. $$
Then
$$ I''(k) = \int_0^{\infty} \frac{-x^2\cos{kx}}{1+x^2} \, dx = I(k) + \int_0^{\infty} \cos{kx} \, dx $$
At this point you should panic and say that this integral is not defined. Correct, but formally it is $2\pi\delta(k)$. Since we are only dealing with positive $k$, I'm going to forget about it (which, again, is illegal; perhaps the more sensible way to do it is to insert a factor $e^{-\lambda x}$, get the last answer as
$$ \int_0^{\infty} e^{-\lambda x} \cos{kx} \, dx= \frac{\lambda}{\lambda^2+k^2}, $$
and then set $\lambda=0$.) The solution to the differential equation
$$ I''(k)-I(k)=0 $$
is $I(k)=A e^k + Be^{-k}$, but the Riemann-Lebesgue lemma (or doing some integration by parts on the original integral) tells us that $A=0$. Therefore we only have to find $B=I(0)$, which is also known as
$$ \int_0^{\infty} \frac{dx}{1+x^2} = \frac{\pi}{2}, $$
which gives the answer.

*Actually do the integral (I): okay, now we're getting serious. Notice that
$$ \int_0^{\infty} f(x) \cos{kx} \, dx = \frac{1}{2}\int_0^{\infty} f(x) e^{ikx} \, dx + \frac{1}{2}\int_0^{\infty} f(x) e^{-ikx} \, dx = \frac{1}{2}\int_{-\infty}^{\infty} (f(x)+f(-x))e^{ikx} \, dx, $$
so this gives in the case of our even function,
$$ I(k) = \int_{-\infty}^{\infty} \frac{e^{ikx}}{1+x^2} \, dx, $$
and we now do the integral using complex analysis, taking a closed semicircular contour in the upper half-plane, using Jordan's lemma to show the integral over the semicircle vanishes, and finding the residue at $x=i$.

*Actually do the integral (II): fine, you say, but what about without complex analysis? Use the Schwinger parametrisation,
$$ \frac{1}{1+x^2} = \int_0^{\infty} e^{-\alpha(1+x^2)} \, d\alpha, $$
and interchange the order of integration. You then have to do
$$ \int_0^{\infty} e^{-\alpha x^2} \cos{kx} \, dx, $$
which can be done by differentiating with respect to $k$ and integrating by parts. Finally, do the $\alpha$ integral.
A: Let $I(k)$ denote the given integral. Take the Laplace transform:
$$\begin{align*}\mathcal{L}\{I(k)\}&=\int_0^\infty\int_0^\infty \frac{\cos kx}{1+x^2}e^{-sk}\,dk\,dx\\\\
&=\int_0^\infty\frac{1}{1+x^2}\mathcal{L}\{\cos kx\}\,dx\\\\
&=\int_0^\infty\frac{s}{(1+x^2)(s^2+x^2)}\,dx\\\\
&=\int_0^\infty\left(\frac{s}{(s^2-1)(1+x^2)}-\frac{s}{(s^2-1)(s^2+x^2)}\right)\,dx\\\\
&=\frac{s}{s^2-1}\left(\frac{\pi}{2}-\frac{\pi}{2s}\right)\end{align*}$$
Finding the inverse gives the desired result.
A: Just write the integral as
$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos kx}{1+x^2} = \frac{1}{4}\int_{-\infty}^{\infty} \frac{e^{ikx}}{1+x^2}dx + \frac{1}{4}\int_{-\infty}^{\infty} \frac{e^{-ikx}}{1+x^2}dx $$
and now check the Fourier transform and inverse Fourier transform of $\frac{1}{1+x^2}$
