$$$$
$$
\int_0^{\infty} \frac{\sin x - x}{x^2 e^x} dx~=~?
$$
解説
$$\begin{align*}
&\int_0^{\infty} \frac{\sin x - x}{x^2 e^x} dx \\
&= \int_0^{\infty} \frac{1}{x^2 e^x} \int_0^1 \frac{d}{dt} (\sin xt - xt ) dt dx \\
&= \int_0^{\infty} \int_0^1 \frac{\cos xt - 1}{x e^x}dt dx\\
&= \int_0^{1} \int_0^{\infty} \frac{1}{xe^x} \int_0^t \frac{d}{du} \cos xu du dx dt\\
&= -\int_0^1 \int_0^{\infty} \int_0^t e^{-x}\sin xu du dx dt\\
&= -\int_0^1 \int_0^t \int_0^{\infty} e^{-x}\sin xu dx du dt\\
&= -\int_0^1 \int_0^t \frac{u}{1+u^2} du dt&(部分積分)\\
&= -\frac12 \int_0^1 \int_0^t \frac{(1+u^2)'}{1+u^2} du dt\\
&= -\frac12 \int_0^1 \ln (1+t^2) dt \\
&= -\frac12 \Biggr[ t\ln(1+t^2) \Biggl]_0^1 + \frac12 \int_0^1 \frac{t}{1+t^2}\cdot 2t dt\\
&= -\frac{\ln 2}{2} + \int_0^1 \left( 1 - \frac{1}{1+t^2}\right) dt\\
&= -\frac{\ln 2}{2} + 1 - \Biggr[ \arctan t\Biggl]_0^1\\
&= 1 - \frac{\pi}{4} - \frac{\ln 2}{2}
\end{align*}$$