My question is:

Is it possible to compute the integral $$\int_{0}^{\infty} \frac{\cos x}{1+x^2} \mathrm{d}x$$ using ODE?

My trial: Let $$ I(a,b) = \int_{0}^{\infty} e^{-bx}\frac{\cos ax}{1+x^2} \mathrm{d}x $$ then by Dominant Convergence theorem, $I(a,b)$ is continuous on $[0,2] \times [0,1]$. So we only need to compute $I(1,0)$. Fix any $b\in (0,1]$, we can get the following ODE: $$ I(a,b)-I^{''}_{aa}(a,b) = \int_{0}^{\infty} e^{-bx}\cos ax \mathrm{d}x=\frac{b}{a^2 + b^2} $$ I have difficulty to proceed. It seems hard to solve this second order ODE. Or any other method using ODE to compute this?

Thank you!


Hint. By setting $$ f(s):=\int_{-\infty}^\infty \frac{s\cos x}{s^2+x^2}\:dx, \qquad s>0, $$ one may prove that $$ f''(s)=\int_{-\infty}^\infty \frac{\partial^2}{\partial s^2}\left(\frac{s\cos x}{s^2+x^2}\right)dx=\int_{-\infty}^\infty \frac{s\cos x}{s^2+x^2}\:dx=f(s) $$ where we have used integration by parts twice. Thus, by using a standard solution of the linear ODE, $$ y''(s)=y(s) $$ one gets$$ y(s)=c_1e^s+c_2e^{-s} $$ then one ends up with

$$ \int_{0}^\infty \frac{s\cos x}{s^2+x^2}\:dx=\frac \pi2 e^{-s},\qquad s>0. $$

The sought integral is obtained by putting $s=1.$

  • $\begingroup$ how does one solve for $c_1,c_2$. One value of s could be $0$ what should be the other value $\endgroup$ – Piyush Divyanakar Dec 12 '17 at 6:08
  • 2
    $\begingroup$ @PiyushDivyanakar Let $s \to \infty$, it gives $c_1=0$, then use $\int_{0}^\infty \frac{1}{1+x^2}\:dx=\frac \pi2$ to obtain $c_2$. $\endgroup$ – Olivier Oloa Dec 12 '17 at 6:11
  • $\begingroup$ @OlivierOloa How to use integrate by part here? Thank you! $\endgroup$ – Edward Wang Dec 13 '17 at 13:20
  • $\begingroup$ @EdwardWang I'm really busy these days, when I've time I will provide some details. Thank you. $\endgroup$ – Olivier Oloa Dec 13 '17 at 23:22
  • 1
    $\begingroup$ @OlivierOloa Oh I see. Thanks a lot. And wish you happy new year! $\endgroup$ – Edward Wang Dec 28 '17 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.