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