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

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!

## 1 Answer

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 $s=1.$

• how does one solve for $c_1,c_2$. One value of s could be $0$ what should be the other value – Piyush Divyanakar Dec 12 '17 at 6:08
• @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 $c_2$. – Olivier Oloa Dec 12 '17 at 6:11
• @OlivierOloa How to use integrate by part here? Thank you! – Edward Wang Dec 13 '17 at 13:20
• @EdwardWang I'm really busy these days, when I've time I will provide some details. Thank you. – Olivier Oloa Dec 13 '17 at 23:22
• @OlivierOloa Oh I see. Thanks a lot. And wish you happy new year! – Edward Wang Dec 28 '17 at 18:03