KVPY 2015 #3

$\begin{aligned} \int_0^\infty e^{-t}|x-t|dt & = \int_0^x e^{-t}(x-t) dt + \int_x^\infty e^{-t}(t-x) dt \\ & = x \left( \int_0^x e^{-t} dt - \int_x^\infty e^{-t} dt \right) \color{#3D99F6}{- \int_0^x te^{-t} dt + \int_x^\infty te^{-t} dt} \quad \small \color{#3D99F6}{\text{By integration by parts (see note)}} \\ & = x\left(\left[-e^{-t}\right]_0^x - \left[-e^{-t}\right]_x^\infty \right) \color{#3D99F6}{+ \left[e^{-t}(t+1)\right]_0^x - \left[e^{-t}(t+1)\right]_x^\infty} \\ & = x \left(-e^{-x}+1+0-e^{-x} \right) + (x+1)e^{-x} - 1 + 0 +(x+1)e^{-x} \\ & = \boxed{x + 2e^{-x} - 1} \end{aligned}$

$\color{#3D99F6}{Note:}$

$\begin{aligned} \quad \int te^{-t}dt & = -te^{-t} + \int e^{-t}dt \\ & = -e^{-t}(t+1) + \color{grey}{constant} \end{aligned}$