And now, the answer to last week’s challenge: here are two solutions, one using the polylogarithm, and one without.
Solution 1 (polylogarithm):
We integrate directly in y:
We showed here a way to find that . So now, we need to find . Making the substitution u=-x, we get
This is a polylogarithm, specifically the dilogarithm:
Now we can find from the series definition of the polylogarithm the special cases and , so , and . Thus , and so
Note that our region of integration is the semicircle that forms the upper half of the unit circle. We then convert to polar coordinates: we have 0≤r≤1, 0≤θ≤π, , and as
then our integral becomes:
Now, let . Then , u=1 when θ=0, and u=-1 when θ=π. Then our integral becomes
Now, we could integrate directly on r to get
and this integral gives a dilogarithm:
So, instead, let , so that ; v=0 when r=1, and v→∞ when r=0. Thus, our integral becomes
And we found here that this equals .