Is there anyway we can calculate this integral $\int_{x-1}^{x+1} f(y)e^{\frac{1}{(x-y)^2-1}} \,dy.$ what I want is an explicit function of $x$.
here $f(y)$ is a piecewise continuous function.
$f(y)=1$ when $y\geq0$
$f(y)=0$ when $y<0$
I tried the following,
Set $u=x-y$ then the integral transformed to the following.
$\int_{-1}^{1} f(x-u)e^{\frac{1}{u^2-1}} \,du.$=$\int_{-1}^{x} f(x-u)e^{\frac{1}{u^2-1}} \,du + \int_{x}^{1} f(x-u)e^{\frac{1}{u^2-1}} \,du.$
since $f(x-u)=0$ when $u>x$,
we have the following. $\int_{-1}^{1} f(x-u)e^{\frac{1}{u^2-1}} \,du.$=$\int_{-1}^{x} f(x-u)e^{\frac{1}{u^2-1}} \,du.$
$\int_{-1}^{x} f(x-u)e^{\frac{1}{u^2-1}} \,du.=\int_{-1}^{x}e^{\frac{1}{u^2-1}} \,du.$
Are theses steps valid so far? How do I continue from here.
What I know is $\int_{-\infty}^{\infty} e^{\frac{1}{u^2-1}} \,du=1$.