## Monday Math 35

In a previous post, we found that . What, then, is ?

Previously, we found that for , . Similarly, as , we see that with , . Setting , we see , for .

Thus
.

As we did previously, we solve the integral in each term with the substitution , which gives us:
.

Plugging into our series,
.

We should recognize that last sum as the Dirichlet eta function. Thus, just as , we see .