Monday Math 80

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:

So then

Now we can find from the series definition of the polylogarithm the special cases and , so , and . Thus , and so

Solution 2:
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 .

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