Question

all parts please!

4. The zeta function (8) = 2n=ln,s > 1, plays an important role in many areas of math- ematics, especially number theory (it can also be defined when s is a complex number). In 1736 Leonard Euler was able to prove that 72 (2) = n2 6 1 n=1 In this problem, your will prove this fact using what you know about double integrals and change of variables (the original proof used a different approach). (a) The double integral So' S dx dy exists as an improper integral and may be de- fined as the limit lime +1- So So 1-ry dx dy. By expressing the integrand as a geometric series, show that 1-ry 1 1 I= dx dy = 1 n2 5(2) 1- 0 n=1 (You may freely interchange the integral and sum.) (b) Find and sketch the region S in the uv-plane obtained by applying the following trans- formation T to the square [0, 1] x [0, 1]: T = U-V V2 y = u + v V2 (c) Using (a) and (b), express the integral I as a sum of two double integrals 11 ans 12. We denote by I, the integral over the part of S where 0 < x < V2/2. (d) Using the identity V2 – (V2 sin 0)2 = V2cos @ when cos 0 2 0, evaluate 11. (2) Using the identity arctan( 22sine) = }(3 7 – 0), evaluate In. (f) Conclude that 5(2) =

Answer 1

a) Remember that, using a geometric series, we have $\frac{1}{1-xy} = \sum_{n=0}^\infty (xy)^n$ for . Therefore,

fх — I fiр тр І І

$= \int_0^1 \int_0^1 \sum_{n=0}^\infty (xy)^n \,\text dx\,\text dy$
$= \sum_{n=0}^\infty \int_0^1 \int_0^1 x^n y^n \,\text dx\,\text dy$
$= \sum_{n=0}^\infty \int_0^1 \left( \frac1{n+1} \right )y^n \,\text dy$
$= \sum_{n=0}^\infty \left( \frac1{n+1} \right ) \left( \frac1{n+1} \right )$
$= \sum_{n=0}^\infty \frac{1}{(n+1)^2}$
$= \sum_{n=1}^\infty \frac{1}{n^2}$ .

b) Using the substitution, and the integral becomes,

2 (7R - ) = fic

$I =\iint_{D} \frac{2}{2-u^2 +v^2} \,\text dv\,\text du$

where is the region

06 0.4 0.2 0.2 0.4 0.6 0.8 1 1.2 -0.2 -0.4 -0.6

c)

$I =\iint_{D} \frac{2}{2-u^2+v^2} \,\text dv\,\text du$
$= \int_0^{\sqrt{2}/2} \int_{-u}^u \frac{2}{2-u^2 +v^2}\,\text dv\,\text du+ \int_{\sqrt{2}/2}^{\sqrt 2} \int_{-u}^u \frac{2}{2-u^2 + v^2}\,\text dv\,\text du$
.