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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!

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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 $s=1.$

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @OlivierOloa Oh I see. Thanks a lot. And wish you happy new year! $\endgroup$ – Edward Wang Dec 28 '17 at 18:03

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