Challenge Math Challenge - May 2020

[benorin]

For a second order DE (which was necessarily of that order to get cosine to be cosine again) there will be two solutions (one for each root of the characteristic equation) with two constants (of integration if you will) that need to be solved for by initial conditions, usually either a pair of function values or one of these and the value of the derivative at a given point. I have the value of the function at zero, but to fully determine the solution I need another such value.

I’m always down to learn a new method of calculating problematic integrals: teach me something!

 

[fresh_42]

Spoiler

For $\alpha=0$ we have $$\int_{-\infty}^{+\infty} \frac{dx}{1+x^2}=[\arctan (x)]_{-\infty}^{+\infty} = \dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)=\pi$$
so we may assume $\alpha> 0$ now, substitute $t=\alpha x , \beta= \alpha^{-1}$ and observe using integration by parts twice
\begin{align*}
\int_0^\infty e^{-t}&\sin(\beta t)\,dt = \left. -e^{-t}\sin(\beta t)\right|_0^\infty + \beta \int_0^\infty e^{-t}\cos(\beta t)\,dt\\
&= 0 + \beta\left(\left[ -e^{-t}\cos(\beta t) \right]_0^\infty -\beta \int_0^\infty (-e^{-t})(-\sin(\beta t)) \,dt \right)\\
&= \beta -\beta^2 \int_0^\infty e^{-t}\sin(\beta t)\,dt
\end{align*}
and thus
$$
\int_0^\infty e^{-t}\sin(\beta t)\,dt = \dfrac{\beta}{1+\beta^2}
$$
This means
\begin{align*}
\int_{-\infty}^{+\infty}\dfrac{\cos(\alpha x)}{1+x^2}\,dx &= \int_{-\infty}^{+\infty}\dfrac{\cos(\alpha x)}{x} \cdot \dfrac{x}{1+x^2} \,dx\\
&=2\int_0^\infty \left( \dfrac{\cos(\alpha x)}{x} \int_0^\infty e^{-t}\sin(x t)\,dt \right)\,dx \\
&= \int_0^\infty \int_0^\infty e^{-t} \cdot \dfrac{2\sin(xt)\cos(\alpha x)}{x}\,dt\,dx \\
&= \int_0^\infty \int_0^\infty e^{-t}\cdot \dfrac{\sin(x(t+\alpha))+\sin(x(t-\alpha))}{x}\,dx\,dt \\
&=\int_0^\infty e^{-t} \cdot \left( \dfrac{\pi}{2}\operatorname{sgn}(t+\alpha) + \dfrac{\pi}{2}\operatorname{sgn}(t-\alpha) \right)\,dt\\
&=\pi \int_\alpha^\infty e^{-t}\,dt\\
&= \pi \cdot e^{-\alpha}
\end{align*}