Twisted One 151's Weblog

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.