Derive Eulers famous result that was mentioned in Section 9.3,by completing each
Chapter 0, Problem 3(choose chapter or problem)
Derive Euler’s famous result that was mentioned in Section 9.3,
\(\sum _{n=1} ^{\infty}\ \frac{1}{n^2}=\frac{\pi^2}{6}\), by completing each step.
(a) Prove that \(\int \frac{dv}{2-u^2+v^2}=\frac{1}{\sqrt{2-u^2}} \arctan \frac{v}{\sqrt{2-u^2}}+C\).
(b) Prove that \(I_1=\int _0 ^{\sqrt{2}/2} \int _{-u} ^u \frac{2}{2-u^2+v^2}\ dv\ du=\frac{\pi^2}{18}\) by using the substitution \(u=\sqrt{2} \sin \theta\).
(c) Prove that
\(I_2=\int _{\sqrt{2}/2} ^{\sqrt{2}} \int _{u-\sqrt{2}} ^{-u+\sqrt{2}} \frac{2}{2-u^2 +v^2}\ dv\ du=4\int _{\pi/6} ^{\pi/2} \arctan \frac{1-\sin \theta}{\cos \theta}\ d\theta\)
by using the substitution \(u=\sqrt{2} \sin \theta\).
(d) Prove that trigonometric identity \(\frac{1-\sin \theta}{\cos \theta}=\tan \left(\frac{(\pi / 2)-\theta}{2}\right) \text {. }\).
(e) Prove that \(I_{2}=\int_{\sqrt{2} / 2}^{\sqrt{2}} \int_{u-\sqrt{2}}^{-u+\sqrt{2}} \frac{2}{2-u^{2}+v^{2}} d v d u=\frac{\pi^{2}}{9}\).
(f) Use the formula for the sum of an infinite geometric series to verify that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y\).
(g) Use the change of variables \(u=\frac{x+y}{\sqrt{2}}\ \text{and}\ v=\frac{y-x}{\sqrt{2}}\) to prove that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y=I_{1}+I_{2}=\frac{\pi^{2}}{6}\).
Text Transcription:
sum _n=1 ^infty 1/n^2 =pi^2 /6
int dv/2-u^2 +v^2 =1/sqrt2-u^2 arctan v/sqrt2-u^2 +C
I_1 = int _0 ^sqrt2 /2 int _-u ^u 2/2-u^2 +v^2 dv du = pi^2 /18
u=sqrt2 sin theta
I_2 = int _sqrt2 /2 ^sqrt2 int _u-sqrt2 ^-u+sqrt2 2/2-u^2 +v^2 dv du = 4 int _pi/6 ^pi/2 arctan 1-sin theta / cos theta d theta
1-sin theta/cos theta = tan ((pi/2)-theta/2)
I_2 = int _sqrt2 /2 ^sqrt2 int _u-sqrt2 ^-u+sqrt2 2/2-u^2 +v^2 dv du =pi^2/9
sum _n=1 ^infty 1/n^2 = int _0 ^1 int _0 ^1 1/1-xy dx dy
u=x+y/sqrt2 and v=y-x/sqrt2
sum _n=1 ^infty 1/n^2 = int _0 ^1 int _0 ^1 1/1-xy dx dy=I_1 + I_2 = pi^2 /6
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