How to solve the double integral of the exponential of an absolute value

Question

I would like to integrate the following:

$$I = \int _0^t d \tau \int _0 ^t d\tau 'e^{-\gamma|\tau-\tau'|} \tag{1}$$

where this integral is defined in the real numbers. In my lecture notes is stated that:

$$\int _0^t d \tau \int _0 ^t d\tau 'e^{-\gamma|\tau-\tau'|} = 2\int _0^t d \tau \int _0 ^\tau d\tau 'e^{-\gamma(\tau-\tau')}=\frac{2}{\gamma}t-\frac{2}{\gamma ^2}\left(1-e^{-\gamma t}\right) \tag{2}$$

however I don't get why the first equality holds true, I don't get why it should be the double.

Anyway: since I didn't get the equality I tried to solve it my own way, like this:

$$\int _0^t d \tau \int _0 ^t d\tau 'e^{-\gamma|\tau-\tau'|}=\int _0^t d \tau \left[\int_0^\tau d\tau ' e^{-\gamma(\tau-\tau')} + \int _\tau ^t d\tau ' e^{\gamma(\tau-\tau')} \right]=\frac{1}{\gamma}\int _0 ^t d \tau \left[-2+e^{-\gamma \tau}+e^{\gamma (\tau -t)}\right]=$$

$$=-\frac{2}{\gamma}t-\frac{2}{\gamma ^2}\left(e^{-\gamma t}-1\right)$$

That is not what I should get according to $(2)$. Indeed I obtained the exact opposite: I am missing a minus sign; however I cannot find it and it's driving me a little crazy... Where did I go wrong? And, bonus question, why the first equality in $(2)$ holds?

Answer 1

Always try to use symmetry. The function $|\tau-\tau'|$ is symmetric about the line $\tau=\tau'$, and so is the region of integration ($0\le\tau,\tau'\le t$). So the integral over the triangle $\{0\le\tau'\le\tau\le t\}$ is equal to the integral over the triangle $\{0\le\tau\le\tau'\le t\}$.

Since you haven't shown the details of your computation, it's difficult to say exactly where you went wrong. I'm guessing that you dropped a negative when you took the antiderivative of $e^{-\gamma\tau'}$.

Answer 2

It seems you made a mistake in the estimation if the inner integral. I am doing things much slower than you did:

$$\begin{align} \int^t_0 e^{-\gamma|\tau-\tau'|}\,d\tau'&=\int^\tau_0e^{-\gamma(\tau-\tau')}\,d\tau'+\int^t_\tau e^{-\gamma(\tau'-\tau)}\,d\tau'\\ &=e^{-\gamma\tau}\Big(\frac{e^{\gamma\tau}-1}{\gamma}\Big)+e^{\gamma\tau}\Big(\frac{e^{-\gamma\tau}-e^{-\gamma t}}{\gamma}\Big)\\ &=\frac1\gamma\big(2-e^{-\gamma\tau} -e^{-\gamma(t-\tau)}\big) \end{align} $$

Answer 3

Let

$$ H(x) = \left\{ \begin{array}{rl} 1, & x > 0 \\ 0, & x\leq 0 \end{array} \right. $$

be a Heaviside function. Then $$ \begin{aligned} I &= \int_0^td\tau\int_0^td\tau'e^{-\gamma|\tau-\tau'|} = \\ &= \int_0^t\left[\int_0^t\left(H(\tau-\tau')e^{-\gamma(\tau-\tau')}+H(\tau'-\tau)e^{-\gamma(\tau'-\tau)}\right)d\tau'\right]d\tau = \\ &= \int_0^t\left[\int_0^\tau e^{-\gamma(\tau-\tau')}d\tau'+\int_\tau^t e^{-\gamma(\tau'-\tau)}d\tau'\right]d\tau = \\ &= \int_0^t\left[e^{-\gamma\tau}\cdot\left.\frac{1}{\gamma}e^{\gamma\tau'}\right|_0^\tau+e^{\gamma\tau}\cdot\left.\left(-\frac{1}{\gamma}\right)e^{-\gamma\tau'}\right|_\tau^t\right]d\tau = \\ &= \int_0^t\left[\frac{1-e^{-\gamma\tau}}{\gamma}+\frac{1-e^{-\gamma t}e^{\gamma\tau}}{\gamma}\right]d\tau = \\ &= \int_0^t\left[\frac{2}{\gamma}-\frac{1}{\gamma}e^{-\gamma\tau}-\frac{e^{-\gamma t}}{\gamma}e^{\gamma\tau}\right]d\tau = \\ &= \left.\frac{2}{\gamma}\tau\right|_0^t + \left.\frac{1}{\gamma^2}e^{-\gamma\tau}\right|_0^t - \left.\frac{e^{-\gamma t}}{\gamma^2}e^{\gamma\tau}\right|_0^t = \\ &= \frac{2t}{\gamma}+\frac{e^{-\gamma t}-1}{\gamma^2}-\frac{1-e^{-\gamma t}}{\gamma^2} = \frac{2}{\gamma}t - \frac{2}{\gamma^2}(1-e^{-\gamma t}) \end{aligned} $$