Convolutions : find $f*g(x)=\int_{R}f(x-y)g(y)dy$

Question

Today I need calculate this convolution :

Given :

$f(x)=e^{x}1_{]-a,a[}(x)$ and $g(x)=e^{-x}1_{]-b,b[}(x)$

Where : $a<b$ , $a,b\in R$

Then find : $f*g(x)=\int_{R}f(x-y)g(y)dy$

My try :

Call : $J=f*g(x)=\int_{R}f(x-y)g(y)dy$

$=\int_0^{t}e^{x-y}1_{]-a,a[}(x-y).e^{-y}1_{]-b,b[}(y)dy$

$=e^{x}\int_{]-b,b[}e^{-2y}1_{]x-a,x+a[}(y)dy$

$=e^{x}\int_{]-b,b[∩]x-a,x+a[}e^{-2y}dy$

Now call $D=]-b,b[∩]x-a,x+a[$ then decision the

intersection :

$D=]x-a,x+a[$ if $|x|≤b-a|$

$D=]-b,x+a[$ if $x+a>-b$ and $x-a<-b$ so : $-b-a<x<-b+a$

$D=]x-a,b[$ if $x+a>b$ and $x-a<b$ so $b-a<x<b+a$

$D=\{\varnothing \}$ if : $x+a>b$ and $x-a<-b$ so $b-a<x<a-b$

Then just do integration!

Is my work true and complete i mean intersection ???

Answer 1

Here is my attempt; note that I use $(-a,a)$ where you would use $]-a,a[$. Since the functions in the convolution are supported in $[-b,\infty)$, we can set $-b$ as the lower integration limit and $x$ as the upper, which results in a proper integral: $$\begin{eqnarray*} f\star g(x) &=& \int_{-b}^x f(x-y)g(y)dy = \int_{-b}^x e^{x-y}1_{(-a,a)}(x-y) e^{-y} 1_{(-b,b)} (y)dy\\ &=& \int_{-b}^x e^{x-2y }1_{(-a,a)}(x-y)1_{(-b,b)} (y)dy\\ &=& \left[\frac{-1}{2}e^{x-2y}( 1_{(-a,a)}(x-y) + 1_{(-b,b)}(y))\right]_ {y=-b}^{y=x}\\ &=& \left[\frac{-1}{2}e^{x-2x}( 1_{(-a,a)}(x-x) + 1_{(-b,b)}(x))\right]\\ &&- \left[\frac{-1}{2}e^{x-2(-b)}( 1_{(-a,a)}(x-(-b)) + 1_{(-b,b)}(-b))\right]\\ &=& \left[\frac{-1}{2}e^{-x}( 1_{(-a,a)}(0) + 1_{(-b,b)}(x))\right] - \left[\frac{-1}{2}e^{x+2b}( 1_{(-a,a)}(x+b) + 1_{(-b,b)}(-b))\right]\\ &=& \left[\frac{-1}{2}e^{-x}( 1 + 1_{(-b,b)}(x))\right] - \left[\frac{-1}{2}e^{x+2b}( 1_{(-a,a)}(x+b) + 0)\right]\\ &=& \frac{e^{x+b}1_{(-(a+b),a-b)}(x) - e^{-x}(1+1_{(-b,b)(x)})}{2}. \end{eqnarray*} $$

NB: answer edited following @Masacroso's comment, pointing out my mistake in my previous attempt.

Answer 2

Note: to make sense the question it must be the case that $a>0$ also, otherwise $(-a,a)=\emptyset$.

We have that $x-y\in(-a,a)\iff -y\in(-a-x,a-x)\iff y\in(x-a,x+a)$, hence

$$f*g(x)=\int e^{x-y}e^{-y}\chi_{(-a,a)}(x-y)\chi_{(-b,b)}(y)\,dy\\ =e^x\int e^{-2y}\chi_{(x-a,x+a)}(y)\chi_{(-b,b)}(y)\, dy\\ =e^x\int e^{-2y}\chi_{(x-a,x+a)\cap(-b,b)}(y)\, dy\tag1$$

and note that

$$(x-a,x+a)\cap(-b,b)=(\max\{x-a,-b\},\min\{b,x+a\})\\ =\left(\frac{x-(a+b)+|x+(b-a)|}2,\frac{x+(a+b)-|x-(b-a)|}2\right)\tag2$$

Thus

$$f*g(x)=\frac12(e^{-|x+(b-a)|+(a+b)}-e^{|x-(b-a)|-(a+b)})\chi_{(-(b+a),b+a)}(x)\tag3$$

where the last indicator function comes from the fact that if $x\le-(b+a)$ or $x\ge (b+a)$ then the interval on $(2)$ would be empty.