How might one perform the unit-square integral

?

First, let us swap the order of integration:

.

Next, we approach the inner integral by performing the u-substitution

→

The limits become *y*=0 → *u*=∞, *y*=1 → *u*=-ln(*x*).

So, with this substitution, we find , and

for 0<*x*<1, ln(*x*)<0, so -ln(*x*)>0; -ln(*x*)→∞ as *x*→0, and -ln(1)=0, so the integration region can be seen as between and *x*=1, *u*>0. So, reversing the order of the double integral,

We found previously that , and the gamma function is defined as ; this means

This is called Hadjicostas’s formula, and holds for *s* any complex number with .

Note that as the pole of *ζ*(*s*) at *s*=1 is of order one: the Laurent series is of the form ,

where the are called the Stieltjes constants, and , the Euler-Mascheroni constant; the singularity in the term is thus removable, with

Thus we have

.

Similarly, with the related integral

,

we can use the same subtitution and change of order:

.

What if we try instead try the integral

?

Performing the same u-substitution:

.

Reversing the order of the double integration, and performing the inner integral,

We found previously that , so then

(for )

and, in analogy to *I*(*s*), we can also find that.

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Tags: Dirichlet Eta Function, Double Integral, Gamma Function, Hadjicostas's Formula, Math, Monday Math, Riemann Zeta Function

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