Series $\sum_{n=1}^{\infty}(-1)^n\frac{n}{(2n-1)^2}$ I how do I evaluate this integral
$$I=\int_{0}^{\pi/2}\cos^2(x/2)\ln\cot(x/2)dx?$$
I would start out by making a sub of $u=\frac{x}{2}$, $dx=2du$
$$I=2\int_{0}^{\pi/4}\cos^2(u)\ln(\cot u)du$$
We shall use the trig: $\sec^2(u)=\tan^2(u)+1$, we sneakily can write as, $\cos^2(u)=\frac{\sec^2(u)}{[\tan^2(u)+1]^2}$
$$I=-\int_{0}^{\pi/4}\sec^2(u)\frac{\ln(\tan u)}{[1+\tan^2(u)]^2}du$$
We are going to make another sub: $v=\tan(u)$, $du=\frac{1}{\sec^2(u)}dv$
$$I=-\int_{0}^{1}\frac{\ln(v)}{(1+v^2)^2}dv$$
The denominator $(1+v^2)^{-2}=1-2v^2+3v^4-4v^6+5v^8-\cdots$, I am sure it is permissible to this series
$$I=-\int_{0}^{1}(1-2v^2+3v^4-4v^6+5v^8-\cdots)\ln(v)dv$$
$$I=\sum_{n=1}^{\infty}(-1)^n{n}\int_{0}^{1}v^{2n-2}\ln(v)dv$$
This can easily evalute by using integration by parts
$$J=\int_{0}^{1}v^{2n-2}\ln(v)dv=-\frac{1}{(2n-1)^2}$$
Finally we got to:
$$I=\sum_{n=1}^{\infty}(-1)^n\frac{n}{(2n-1)^2}$$
I only know that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^2}=G$$
Where $G$ represent the Catalan's Constant, approx to $0.91596...$
 A: Define
$$f(x)=\sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{2n-1}=-x\arctan(x)$$
Then differentiate this to obtain
$$f'(x)=\sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n-1}}{2n-1}=-\arctan(x)-\frac{x}{x^2+1}$$
Divide both sides by $x$ to obtain
$$\frac{f'(x)}{x}=\sum_{n=1}^\infty \frac{(-1)^n 2n x^{2n-2}}{2n-1}=-\frac{\arctan(x)}{x}-\frac{1}{x^2+1}$$
Then integrate both sides from $x=0$ to $1$, giving
$$\int_0^1\frac{f'(x)}{x}dx =\sum_{n=1}^\infty \frac{(-1)^n 2n}{(2n-1)^2}=-\int_0^1\frac{\arctan(x)}{x}dx-\int_0^1\frac{dx}{x^2+1}$$
The first integral on the RHS is equal to $G$, the Catalan Constant, which can be shown by expanding the arctangent in the integrand with its Taylor Series. The second integrand has an elementary antiderivative. Thus, we have
$$\int_0^1\frac{f'(x)}{x}dx =\sum_{n=1}^\infty \frac{(-1)^n 2n}{(2n-1)^2}=-G-\frac{\pi}{4}$$
or
$$\sum_{n=1}^\infty \frac{(-1)^n n}{(2n-1)^2}=-\frac{G}{2}-\frac{\pi}{8}$$
A: The sum is easily evaluated by recognizing that
$$\frac{n}{(2 n-1)^2} = \frac12 \frac1{2 n-1} + \frac12 \frac1{(2 n-1)^2} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}\pars{-1}^{n}{n \over \pars{2n - 1}^{2}}}  =
{1 \over 2}\,\ic\sum_{n = 1}^{\infty}\ic^{2n - 1}\,\,{\pars{2n - 1} + 1
\over \pars{2n - 1}^{2}}
\\[5mm] = &\
{1 \over 2}\,\ic\sum_{n = 1}^{\infty}\ic^{n}\,{n + 1
\over n^{2}}\,{1^{n} - \pars{-1}^{n} \over 2}
\\[5mm] = &\
-\,{1 \over 2}\,\Im\pars{\sum_{n = 1}^{\infty}{\ic^{n} \over n} +
\sum_{n = 1}^{\infty}{\ic^{n} \over n^{2}}}
\\[5mm] = &\
{1 \over 2}\,\
\underbrace{\Im\ln\pars{1 - \ic}}_{\ds{-\,{\pi \over 4}}}\ -\ {1 \over 2}\
\underbrace{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over
\pars{2n + 1}^{2}}}_{\ds{G}}
\\[5mm] = &\
\bbx{-\,{\pi \over 8} - {1 \over 2}\,G} \approx -0.8507\\ &
\end{align}
$\ds{G}$ is the Catalan Constant.
A: $$\sum_{n\geq 1 } (-1)^n \frac{n}{(2n-1)^2} =\frac{1}{2}\sum_{n\geq 1}(-1)^n\frac{(2n-1)+1}{(2n-1)^2}$$ since we have the convergent series so$$\frac{1}{2}\sum_{n\geq 1}\left(\frac{(-1)^n}{2n-1}+\frac{(-1)^n}{(2n-1)^2}\right)=\frac{1}{2}\left(-\frac{\pi}{4}-G\right)=-\frac{\pi}{8}-\frac{G}{2}$$
Here is $\sum_{n\geq 1} \frac{(-1)^n}{2n-1}$ is well know series called Leibniz formula for $\pi$ that converges to $-\frac{\pi}{4}$ . $$\sum_{n\geq 1} \frac{(-1)^n}{2n-1} =\sum_{n\geq 0}\frac{(-1)^{n+1} }{2n+1} =\sum_{n\geq 0}\left(-\frac{1}{4n+1}+\frac{1}{4n+3}\right)=\frac{1}{4}\left(-\psi^0\left(\frac{3}{4}\right)+\psi^{0}\left(\frac{1}{4}\right)\right)=-\frac{\pi}{4}$$
A: To evaluate the integral without resorting to the series sum
$$\int_{0}^{\pi/2}\cos^2\frac x2\ln\cot \frac x2dx 
\overset{v=\tan\frac x2}=-\int_{0}^{1}\frac{2\ln vdv}{(1+v^2)^2}
=- \int_{0}^{1}\frac{\ln v}{v}d\left(\frac{v^2}{1+v^2}\right)dv\\
\overset{IBP}= \int_{0}^{1} \frac{1}{1+v^2}dv -\int_{0}^{1} \frac{\ln v}{1+v^2}dv
=\frac\pi4+G
$$
