Integral of an exponential with abosolute values in the exponent

Question

I would like to know how can the following integral be computed

$\int_{-\infty}^{+\infty}dw'e^{-k\mid w-w'\mid} e^{-k\mid w'\mid}$

All information is really appreciated.

Answer 1

Perhaps I am going nuts, but let's try.

Firs of all, the integrand depends on the sign of both $\omega - \omega'$ and $\omega'$. So we split initially into two cases:

• When $\omega - \omega' > 0$ that is, $\omega' < \omega$ AND $\omega' > 0$

This means that $|\omega - \omega'| = \omega - \omega'$ and $|\omega'| = \omega'$ hence the integral becomes

$$\int_0^{\omega} e^{-k(\omega - \omega')}e^{-k\omega'}\ \text{d}\omega' = \int_0^{\omega} e^{-k\omega}\ \text{d}\omega = \boxed{\frac{1-e^{-k \omega }}{k}}$$

• When $\omega - \omega' > 0$, that is again $\omega' < \omega$ BUT $\omega' < 0$

This means that $|\omega - \omega'| = \omega - \omega'$ and $|\omega'| = -\omega'|$ hence the integral becomes

$$\int_{-\infty}^{\omega} e^{-k(\omega - \omega'}e^{k\omega'}\ \text{d}\omega' = \int_{-\infty}^{+\omega} e^{-k\omega + 2k\omega'}\ \text{d}\omega' = \boxed{\frac{1}{2k} e^{k\omega}}$$

Now we have to face the two other cases, that is:

• When $\omega - \omega' < 0$ that is, $\omega' > \omega$ AND $\omega' > 0$

• When $\omega - \omega' < 0$ that is, $\omega' > \omega$ BUT $\omega' < 0$

In the first of the latter cases the integral becomes

$$\int_{\omega}^{+\infty} e^{k(\omega - \omega')}e^{-k\omega'}\ \text{d}\omega' = e^{k\omega}\int_{\omega}^{+\infty}e^{-2k\omega'}\ \text{d}\omega' = \boxed{\frac{1}{2k}e^{k\omega}}$$

In the last of the later cases the integral becomes

$$\int_{\omega}^{0} e^{k(\omega - \omega')}e^{k\omega'}\ \text{d}\omega' = = \int_{\omega}^0 e^{k\omega}\ \text{d}\omega' = \boxed{\frac{1-e^{k \omega }}{k}}$$