Just wondering if anyone could help (give steps or explanation) on how to give Euler's Demonstration. A question I've come across is:
Give Euler's Demonstration that $$ \sum^{\infty}_{k=1} \frac {1} {(2k-1)^2}=\frac {\pi^2}{8} $$
I think I need the expansion of cos but I'm unsure how to go about showing the above. Any help would be greatly appreciated!
This gives the approach in Aigner and Ziegler described earlier. First, define an integral $I$ and show it is equal to the sum over $k$ of $1/(2k+1)^2$:
$$ \begin{aligned} I = \int_0^1 \int_0^1 \frac{1}{1 - x^2 y^2} dx dy &= \int_0^1 \int_0^1 \sum_{k=0}^\infty (x^2 y^2)^k dx dy \\ &= \int_0^1 \left( \sum_{k=0}^\infty \int_0^1 x^{2k} y^{2k} dx \right) dy \\ &= \int_0^1 \left( \sum_{k=0}^\infty \frac{1}{2k+1} y^{2k} \right) dy \\ &= \sum_{k=0}^\infty \left( \int_0^1 \frac{1}{2k+1} y^{2k} dy \right) \\ &= \sum_{k=0}^\infty \frac{1}{(2k+1)^2} \end{aligned} $$
So we want to find $$ I = \iint\limits_{S} f(x,y) dx dy $$ where $S$ is the unit square $0 < x,y < 1$ and $f = \frac{1}{1 - x^2 y^2}$
Define a substitution $(x,y)$ to $(u,v)$ with the transformation: $$ x = \frac{\sin u}{\cos v} \qquad y = \frac{\sin v}{\cos u} $$
Under the change of variable we have $$ I = \iint\limits_{S} f(x,y) dx dy = \iint\limits_{T} f(x(u,v), y(u,v)) |J| du dv $$ where $J$ is the determinant of Jacobian under the coordinate transformation. $T$ is the image of $S$ in the $uv$ plane under the transformation, which can be shown to be a bijection between $S$ and $T = \{(u,v) : u, v > 0 \, , \, u + v < \pi / 2\}$ (see below).
The Jacobian determinant is $$ J = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial v} & \frac{\partial y}{\partial v} \\ \end{pmatrix} $$ Calculate $J$: $$ J = \det \begin{pmatrix} \frac{ \cos u }{ \cos v } && \frac{ \sin u \sin v }{ \cos^2 v } \\ \frac{ \sin u \sin v }{ \cos^2 u } && \frac{ \cos v }{ \cos u } \\ \end{pmatrix} = 1 - \frac{\sin^2 u \sin^2 v }{\cos^2 u \cos^2 v} $$
So $|J| = 1 - x^2 y^2$ for $0 < x,y < 1$.
This gives $f |J| = 1$ so the integral becomes $$ I = \iint\limits_{T} 1 \; du dv = \int_0^{\pi/2} \int_0^{\pi/2 - v} du dv = area(T) = \frac{\pi^2}{8} $$
So, as required, we have shown that $$ I = \sum_{k=0}^\infty \frac{1}{(2k+1)^2} = \frac{\pi^2}{8} $$
The bijection between $S$ and $T$: The coordinate transformation is described above as a mapping from $T$ to $S$ It is possible to obtain the inverse over $S$
$$ \cos u = \sqrt{\frac{1-x^2}{1 - x^2 y^2}} \qquad \cos v = \sqrt{ \frac{1-y^2}{1-x^2 y^2} } $$
In both directions, it is possible to show that the mappings are 1-1 and that each covers its target region. It is also possible to show that $$ \sin(u + v) = \frac{x+y}{1+xy} $$ which helps to determine how the 'top' and 'right' edges of the unit square are together mapped to the segment of the line $u+v = \pi / 2$ between $(0,\pi/2)$ and $(\pi/2,0)$ in the $uv$ plane.