Alan Smith

2021-12-26

I want to find an expression for

My initial thought was to use the known

zesponderyd

Beginner2021-12-27Added 41 answers

Let $f\left(x\right)=\mathrm{arctan}\left(\frac{x+1}{x-1}\right),\text{}\text{}\mathrm{\forall}\left|x\right|1$

It can be easily showed that:

$f}^{\prime}\left(x\right)=\frac{1}{1+{x}^{2}}=\sum _{n=0}^{\mathrm{\infty}}{(-1)}^{n}{x}^{2n$

Integrating both sides yields that$\mathrm{\exists}C\in \mathbb{R}$ such that:

$\mathrm{arctan}\left(\frac{1+x}{1-x}\right)+C=\sum _{n=0}^{\mathrm{\infty}}\frac{{(-1)}^{n}{x}^{2n+1}}{2n+1}$

Check for f(0) to conclude C and you're done.

It can be easily showed that:

Integrating both sides yields that

Check for f(0) to conclude C and you're done.

Annie Levasseur

Beginner2021-12-28Added 30 answers

Use the identity $\mathrm{arctan}\left(\frac{x+1}{x-1}\right)=-\mathrm{arctan}\left(\frac{x+1}{1-x}\right)=-(\frac{\pi}{4}+\mathrm{arctan}\left(x\right))$

karton

Expert2022-01-08Added 613 answers

Since

you can express the RHS as a power series, and then integrate the result to get the desired series for your original function

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