An integral that arises in the study of the Riemmann- zeta function....

Question

The integral ∫¹₀ ∫¹₀ 1 / ( 1- x² y² ) d x d y arises in the study of the Riemmann- zeta function.

Use the transformation x = (sin u ) / ( cos v ) and y = (sin v ) / ( cos u ) to evaluate the integral.

Answer 1

The "trick" with this problem is that is that the transformation x = (sin u ) / ( cos v ) and y = (sin v ) / ( cos u )

transforms the square 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 into the triangle 0 ≤ u ≤ π/2 , 0 ≤ v ≤ π/2− u. This can be done

by transforming each side of the square separately.

Notice that (sin u ) / ( cos v ) =1 suggests sin u = cos v or equivalently sin u = sin ( π /2 − v )

thus u + v = π/2

Then as Dayv O. correctly observed the Jacobian is the reciprocal of the integrant and therefore

the ∫¹₀ ∫¹₀ 1 / ( 1- x² y² ) d x d y = ∫₀^π/2 ∫₀^{π/2 - u} dv du = ∫₀^π/2 ( π/2 −u )du =[ ( π/2) u − u²/2 ] |₀^π/2 =

π²/4 − π²/8 = π²/8

Answer 2

CORRECTION, answer=π²/8, Which makes more sense than previous answer I provided (which was less than 1 and based on limits 0≤v≤π/4 and 0≤≤π/4) . Please see correct limits of this double integration below.the

original short answer, Jacobian is same as integrand recipricol so have answer = double integral dudv.,,,,

CORRECTION

the limits of integation translated from substitution are 0≤v≤π/2 and π/2-v≤u≤π/2 implying first integration

will be with du. Reason ofr limit of u and v. Can show if 0≤x≤1 - when x=1 u= π/2-v (if v=0 which is neccessary for y=0 u approaches as limit , u=π/2) and when x<1 u>π/2-v