sum of arctangent

Solution 1:

Here is a slightly different approach.

Note that $$ \log(1+ix)=\tfrac12\log(1+x^2)+i\tan^{-1}(x)\tag{1} $$ So, applying $(1)$ yields $$ \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right)=\Im\left(\log\left(\prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)\right)\right)\tag{2} $$ where $\alpha =\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$, so that $\alpha^2=-i$.

Furthermore, the sinc function has the following product expansion $$ \prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)=\frac{\sin(\pi\alpha)}{\pi \alpha}\tag{3} $$ We also have $$ \begin{align} \frac{\sin(x+iy)}{x+iy} &=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\ &=\frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}\\ &+\;i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}\tag{4} \end{align} $$ Combining $(2)$, $(3)$, $(4)$, and using the identity $\tan(\pi/4-x)=\frac{1-\tan(x)}{1+\tan(x)}$ yields $$ \begin{align} \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right) &=\Im\left(\log\left(\frac{\sin(\pi\alpha)}{\pi\alpha}\right)\right)\\ &=\tan^{-1}\left(\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}\right)\\ &=\tan^{-1}\left(\frac{\tan(\frac{\pi}{\sqrt{2}})-\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})+\tanh(\frac{\pi}{\sqrt{2}})}\right)\\ &=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right)\tag{5} \end{align} $$

Solution 2:

I don't understand why you have $x$ in the numerator and $y$ in the denominator in some of the upper equations. However, in $[2]$ they're the right way around. The left-hand side is minus the sum of the arguments of $1-z^2/k^2$, the first term on the right-hand side is the argument of $z$, and the second term on the right-hand side is minus the argument of $\sin(\pi z)$:

$$ \begin{eqnarray} \sin(\pi z) &=& \frac{\mathrm e^{\mathrm i\pi z}-\mathrm e^{-\mathrm i\pi z}}{2\mathrm i} \\ &=& \frac{\mathrm e^{\mathrm i\pi (x+\mathrm iy)}-\mathrm e^{-\mathrm i\pi (x+\mathrm iy)}}{2\mathrm i} \\ &=& \frac{\mathrm e^{-\pi y}(\cos(\pi x)+\mathrm i\sin(\pi x))-\mathrm e^{\pi y}(\cos(\pi x)-\mathrm i\sin(\pi x))}{2\mathrm i} \\ &=& \sin(\pi x)\cosh (\pi y)+\mathrm i\cos(\pi x)\sinh(\pi y)\;. \end{eqnarray} $$

By the way, a lot of your parentheses are undersized.

Solution 3:

I see now. Thanks for the pointers, Joriki.

The roots of $1-\frac{z^{2}}{k^{2}}$ are $-k$ and $k$.

This gives $$\frac{\frac{x-k}{y}+\frac{x+k}{y}}{1+\frac{x-k}{y}\frac{x+k}{y}}$$

$$=\frac{2xy}{k^{2}-x^{2}+y^{2}}$$.