Find :

1). Using an infinite series and the Riemann zeta function

2). Using neither infinite series nor polylogarithms

1).Infinite Series

The infinite geometric series holds for , which, except for the singularity at (*x*,*y*)=(1,1), satisfies for our region of integration; thus

;

we can exchange the order of integration and summation

and performing the integral , we see

.

Now,

;

we can separate this sum into sums over the odd and even integers

;

But the sum over the even integers is

, and so

and so .

2). Without series or polylogarithms.

If we try to integrate directly over one variable, say *x*, we see that ,

so we obtain single integral

,

and since , we have

,

which requires polylogarithms to integrate.

So instead, we need to make a change of variables. The key is to define new variables *u* and *v*, with and . To require a one-to-one mapping to positive *x* and *y*, we need , . Further, we see *x*≤1 gives us

,

and similarly, the upper limit on *y* gives

.

Thus, we see the region of integration *R* in the *uv* plane is the isoceles right triangle bounded by *u*=0, *v*=0, and .

Now, we need to find the Jacobian determinant, since

.

We find

.

Thus, we have

,

and our integral is just equal to the area of the triangle *R*, which is

.

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Tags: Double Integral, Geometric Series, Integral, Jacobian, Math, Monday Math, Polylogarithm, Riemann Zeta Function

This entry was posted on November 8, 2010 at 12:27 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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