convolution between $f=e^{-x/5}\chi_{[0,\infty)}(x)$ and $g=e^{x/3}\chi_{(-\infty,0]}(x)$

Solution 1:

For the functions like this (described in one relation) it is routine to evaluate convolution actually, beware of multi-relation functions like piece-wise linear ones and other similar functions.

Notice that $f(x) = e^{-\frac{x}{5}}u(x)$ and $g(x) = e^{\frac{x}{3}}u(-x)$ where $u(x)$ is the unit step function aka Heaviside step function which is

$$u(x) = \begin{cases} 1 \qquad,x\ge0\\ 0\qquad, \text{otherwise} \end{cases}$$

Now we write the definition of convolution

$$h(x) = f(x)*g(x) = \int_{-\infty}^{\infty}f(\lambda)g(x-\lambda)d\lambda$$

And now substitute

$$h(x) = \int_{-\infty}^{\infty}e^{-\frac{\lambda}{5}}u(\lambda)e^{\frac{x-\lambda}{3}}u(-(x-\lambda)) d\lambda = \begin{cases} e^{\frac{x}{3}}\int_{0}^{\infty}e^{-\frac{\lambda}{5}}e^{-\frac{\lambda}{3}} d\lambda \qquad \text{if }\; x < 0\\ e^{\frac{x}{3}}\int_{x}^{\infty}e^{-\frac{\lambda}{5}}e^{-\frac{\lambda}{3}} d\lambda \qquad \text{if }\; x \ge 0 \end{cases}$$

Where we used the fact that $u(\lambda)\neq 0$ when $\lambda \ge 0$ and similarly $u(-(x-\lambda))\neq 0$ when $\lambda\ge x$. Now we evaluate the last integrals

$$h(x) = \begin{cases} e^{\frac{x}{3}}\int_{0}^{\infty}e^{-\frac{\lambda}{5}}e^{-\frac{\lambda}{3}} d\lambda = \frac{15}{8}e^{\frac{x}{3}} \qquad \text{if } \; x < 0\\ e^{\frac{x}{3}}\int_{x}^{\infty}e^{-\frac{\lambda}{5}}e^{-\frac{\lambda}{3}} d\lambda = \frac{15}{8}e^{-\frac{x}{5}} \quad\;\; \text{if }\; x \ge 0 \end{cases}$$

Therefore $h(x) = \frac{15}{8}e^{\frac{x}{3}}u(-x) + \frac{15}{8}e^{-\frac{x}{5}}u(x)$

Now you try to calculate this integral like $h(x) = g(x)*f(x) = \int_{-\infty}^{\infty}g(\lambda)f(x-\lambda)d\lambda$

and see that you get the same result.