How to integrate $\int_0^\infty e^{-t}|x-t|dt$?

I tried writing it as: $\int_0^x e^{-t}(x-t)dt$ $-\int_x^\infty e^{-t}(x-t)dt$. How do I proceed further?

  • $\begingroup$ You’ve got the right idea. Just use integration by parts now. But i think the minus sign should be on the right integral. $\endgroup$ – Ryan Greyling Aug 6 at 16:54
  • $\begingroup$ @YuriyS Yes, I just noticed. $\endgroup$ – Tapi Aug 6 at 16:55
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    $\begingroup$ As a hint, can you do the indefinite integral $\int e^{-t} t dt$. As another hint, you can do it using integration by parts $\endgroup$ – Yuriy S Aug 6 at 16:55

You have that

$$\int_0^\infty e^{-t}|x-t|~dt= \int_0^x e^{-t}(x-t)~dt-\int_x^\infty e^{-t}(x-t)~dt$$

so you can further break up the integrals on the RHS by

\begin{align}\int_0^x e^{-t}(x-t)~dt-\int_x^\infty e^{-t}(x-t)~dt &= x\int_0^x e^{-t}~dt-\int_0^x e^{-t}t~dt-x\int_x^\infty e^{-t}~dt + \int_x^\infty e^{-t}t~dt\end{align}


$$x\int_0^x e^{-t}~dt = x\Big(\sinh(x) - \cosh(x) + 1\Big)$$

since $e^{x}=\sinh(x) + \cosh(x) \implies -e^{-x}=\sinh(x) - \cosh(x)$ and

$$x\int_x^{\infty} e^{-t}~dt = x\Big(e^{-x}\Big)$$

because $1/e^{\infty}$ is zero. To evaluate the remaining two integrals, I would apply integration by parts and set $u=t$ and $dv=e^{-t}dt$.


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