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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Tags: Integral, Math, Monday Math, Polylogarithm

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