I am looking for examples of problems that are easier in the infinite case than in the finite case. I really can’t think of any good ones for now, but I’ll be sure to add some when I do.

**Answer**

One can compute the value of

∫∞0e−t2dt

exactly. This is known as the Gaussian integral, and it has its own Wikipedia page. The answer turns out to be 12√π.

But one cannot do the same with

∫x0e−t2dt

because the antiderivative of the integrand is not an elementary function. This is why we gave a name to the error function erf(x)=2√π∫x0e−t2dt, which also has its own Wikipedia page.

In that sense, the infinite case is easier than the finite case.

*Addendum:* The same phenomenon occurs for variants of this integral, in particular we can transform the integrand to evaluate ∫∞−∞ae−(t−b)2/(2c2)dt=√2a|c|√π as detailed here on Wikipedia.

**Attribution***Source : Link , Question Author : Asinomás , Answer Author : Will R*