<s>[INST] expressing $\int tan^{-1}tdt$ as a power series I have a question which asks me to state $\int_0^x\tan^{-1}(t)dt$ as a power series in $x$ and then use that result to show that $\frac{\pi}{4}-\log\sqrt{2}= 1-1/2-1/3+1/4+1/5--++\cdots$
For the first part:
$$\tan^{-1}t= \int\sum_{n=0}^\infty(-1)^nx^{2n}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\cdots$$6
then integrating from $0$ to $x$ we get 
$$\int_0^x\tan^{-1}(t)dt= \frac{x^2}{2}-\frac{x^4}{3\cdot4}+\frac{x^5}{5\cdot6}-\frac{x^8}{7\cdot8}\cdots=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{n(n-1)}$$
But how can I use this to show the second part?
 [/INST] The integral itself is
$$\int_0^1 dt \, \arctan{t} = \frac{\pi}{4} - \log{\sqrt{2}}$$
The series is, upon integration,
$$\sum_{k=1}^{\infty} \frac{(-1)^{n+1}}{2n (2n-1)} = \sum_{k=1}^{\infty} (-1)^{n+1} \left(\frac{1}{2 n-1} - \frac{1}{2 n}\right)$$
Therefore
$$\begin{align}\frac{\pi}{4} - \log{\sqrt{2}} &= \left ( 1-\frac{1}{2}\right ) - \left ( \frac{1}{3}-\frac{1}{4}\right ) + \left ( \frac{1}{5}-\frac{1}{6}\right ) - \left ( \frac{1}{7}-\frac{1}{8}\right ) -\ldots\\ &= 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5}- \frac{1}{6} - \frac{1}{7}+ \frac{1}{8}+\ldots \end{align}$$
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