A combination integral and series resulting the inverse tangent integral

Solution 1:

I stress that this is not a solution, but such a comment was too long to fit above.

I got a different formula for \begin{equation} \int_0^{\Large\frac{\pi}{2}}\sin(2n-1)x\cot x\,dx=\frac{(-1)^{n-1}}{2n-1}+2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1} \end{equation} than what you posted. But after close inspection, they can be shown to be equivalent by induction. Anyway, I put my work here for clarity in case others are curious.

I'd also like to point out the identity \begin{equation*} 2\tan^{-1}(z) = \tan^{-1}\left(\frac{2z}{1-z^{2}}\right) \end{equation*} which may help in putting the right hand side in the correct form.

Let $w = e^{ikx}$ and note that \begin{align*} \sum_{k=1}^{n-1}w^{2k} & = \frac{w^{2n} - w^{2}}{w^{2} - 1}\\ & = \frac{w^{2n-1} - w}{w - w^{-1}}\\ & = \frac{\cos((2n-1)x) + i \sin((2n-1)x) - \cos{x} - i\sin{x}}{2i\sin{x}}. \end{align*}

Taking the real parts, we have \begin{equation*} \sum_{k=1}^{n-1}\cos(2kx) = \frac{\sin((2n-1)x)}{2\sin{x}} - \frac{1}{2}. \end{equation*}

Multiplying both sides by $\cos{x}$ and rearranging yields \begin{equation*} \sin((2n-1)x)\cot(x) = \cos{x} + 2\sum_{k=1}^{n-1}\cos(2kx)\cos{x} \end{equation*}

Now we use the fact that \begin{equation*} 2\cos(2kx)\cos{x} = \cos((2k+1)x) + \cos((2k-1)x) \end{equation*} to get \begin{align*} \int_{0}^{\pi/2} \sin((2n-1)x)\cot{x}\,dx & = \int_{0}^{\pi/2}\cos{x}\,dx + \sum_{k=1}^{n-1}\int_{0}^{\pi/2}\left[\cos((2k+1)x) + \cos((2k-1)x)\right]\,dx\\ & = 1 + \sum_{k=1}^{n-1}\left[ \frac{\sin((2k+1)x)}{2k+1} + \frac{\sin((2k-1)x)}{2k-1} \right]_{x=0}^{\pi/2}\\ & = 1 + \sum_{k=1}^{n-1}\frac{(-1)^{k}}{2k+1} + \frac{(-1)^{k-1}}{2k-1}\\ & = 1 + \sum_{k=1}^{n-1}(-1)^{k}\frac{-2}{(2k+1)(2k-1)}\\ & = 1 + 2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{(2k+1)(2k-1)} \end{align*}