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

Question

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!

Answer 1

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 $s=1.$

Answer 2

Let’s consider the following integral parameterised by $y$. $$ I(y)=\int_{0}^{\infty} \frac{\sin ( xy)}{x\left(1+x^2\right)},\quad \textrm{ where }y\ge 0. $$ Then we find the 1st and 2nd derivatives of $I(y)$ to set up a differential equation. $$ \begin{aligned} & I^{\prime}(y)=\int_0^{\infty} \frac{\cos (x y)}{1+x^2} d x \\ & I^{\prime \prime}(y)=-\int_0^{\infty} \frac{x \sin (x y)}{1+x^2} d x \end{aligned} $$ Combining them yields $$ -I^{\prime \prime}(y)+I(y)=\int_0^{\infty} \frac{\sin (x y)}{x} d x=\frac{\pi}{2} $$ With the initial conditions $I(0)=0, I’(0)=\frac \pi2$ gives the general solution of $$I(y)= \int_{0}^{\infty} \frac{\sin ( xy)}{x\left(1+x^2\right)}=\frac{\pi}{2}\left(1-e^{-y}\right) $$ Differentiating $I(y)$ w.r.t. $y$ yields $$ \boxed{\int_{0}^{\infty} \frac{\cos (x y)}{1+x^2} d x=\frac{\partial}{\partial y}(I (y))=\frac{\pi}{2} e^{-y}\,} $$ In particular, when $y=1$, $$ \boxed{\int_{0}^{\infty} \frac{\cos x}{1+x^2} d x=\frac{\pi}{2e} } $$