P2

[Victor Ivrii]

Find solution $u(x,t)$ to
\begin{align*}
&u_{tt}-4u_{xx}= \frac{8}{x^2+1},\tag{1}\\
&u|_{t=0}=0, \quad u_t|_{t=0}=0.\tag{2}
\end{align*}

Hint: Change order of integration over characteristic triangle

[Jingxuan Zhang]

A direct computation yields:
\begin{align*}
u &=\frac{1}{4}\int_0^t\int_{x-2t+2t'}^{x+2t-2t'} \frac{8}{x'^2+1} \,dx'\,dt'\\
&=\frac{1}{4}\int_{x-2t}^x\int_{0}^{x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx' +\frac{1}{4}\int_x^{x+2t}\int_{0}^{-x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx'\\
&=\int_{x-2t}^x\frac{x'-x+2t}{x'^2+1} \,dx' + \int_x^{x+2t}\frac{x'+x+2t}{x'^2+1} \,dx' \\
&= \int_{x-2t}^x\frac{x'}{x'^2+1} \,dx + (x-2t) \int_x^{x-2t}\frac{1}{x'^2+1}\,dx'+
\int_x^{x+2t}\frac{x'}{x'^2+1} \,dx + (x+2t) \int_x^{x+2t}\frac{1}{x'^2+1}\,dx'\\
&=\ln(x^2+1)-2x\tan^{-1}(x)-\frac{1}{2}\ln((x-2t)^2+1)+(x-2t)\tan^{-1}(x-2t)-\frac{1}{2}\ln((x+2t)^2+1)+(x+2t)\tan^{-1}(x+2t)
\end{align*}

To my great shame, I failed to directly compute this during the actual sitting.

[Victor Ivrii]

Definitely change of variables makes calculations easier; but one could  make calculations without it. To do so one needs to know the primitive of $]\arctan(x)$:
$$
\int \arctan(x)\,dx = x\arctan(x)-\int x\,d\arctan(x)= x\arctan(x) -\int \frac{x\,dx}{x^2+1}= x\arctan(x) -\frac{1}{2}\ln (x^2+1).
$$