expressing $\int tan^{-1}tdt$ as a power series

Question

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?

Answer 1

You can use the integration-by-parts formula to get the second: \begin{eqnarray*} \int_0^x\tan^{-1}tdt&=&t\tan^{-1}t\big|_0^x-\int_0^x\frac{t}{t^2+1}dt\\ &=&t\tan^{-1}t\big|_0^x-\frac{1}{2}\ln(t^2+1)\big|_0^x\\ &=&x\tan^{-1}x-\frac{1}{2}\ln(x^2+1). \end{eqnarray*} So $$ \int_0^1\tan^{-1}tdt=\frac{\pi}{4}-\frac{1}{2}\ln 2=\cdots. $$

Answer 2

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

$$\begin{align}\frac{\pi}{4} - \log{\sqrt{2}} &= \left ( 1-\frac{1}{2}\right ) - \left ( \frac{1}{3}-\frac{1}{4}\right ) + \left ( \frac{1}{5}-\frac{1}{6}\right ) - \left ( \frac{1}{7}-\frac{1}{8}\right ) -\ldots\\ &= 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5}- \frac{1}{6} - \frac{1}{7}+ \frac{1}{8}+\ldots \end{align}$$