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


marked as duplicate by Gabriel Romon, Lucian, Lord_Farin, kobe, user147263 Mar 26 '15 at 0:12

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  • 1
    $\begingroup$ Compute residues? $\endgroup$ – abnry Mar 25 '15 at 20:03
  • 2
    $\begingroup$ The integral is yearning for parameter differentiation. $\endgroup$ – Gabriel Romon Mar 25 '15 at 20:12
  • 1
    $\begingroup$ Heuristically, if you let $f(k):=\int_0 ^\infty \frac{1}{1+x^2} \cos(kx)$, differentiation under the integral sign yields $f''(k)=-\int_0 ^\infty \frac{x^2}{1+x^2} \cos(kx)$. Hence $f''(k)-f(x)\approx-\int_0^\infty cos(kx)\approx0$. This a nice ODE. $\endgroup$ – Gabriel Romon Mar 25 '15 at 20:21
  • $\begingroup$ Other duplicates can also be found here. $\endgroup$ – Lucian Mar 25 '15 at 20:36

(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:

  1. 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 $1/(1+x^2)$.)

  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.

  3. 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 $x=i$.

  4. 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.


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.

  • $\begingroup$ Of course, that last step's a lot easier when you see that the bracket contains $\frac{s-1}{s}$, so the Laplace Transform is just $\frac{\pi}{2}\frac{1}{1+s}$. $\endgroup$ – Chappers Mar 26 '15 at 3:55

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 $\frac{1}{1+x^2}$


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