The inverse function of $\tan x$ is called $\arctan x$ .
That is, if $\tan \theta =x$ , then $\arctan x=\theta$ .
We have already gotten
\[ \int_{-\infty}^{\infty} \frac{1}{1+x^2}dx =\pi . \]
By some calculations. we will get
\[ \int_0^1 \frac{1}{1+x^2}dx = \frac{\pi}{4} . \]
In additions,
\[ \arctan y =\int_0^y \frac{1}{1+x^2}dx \]
Thus,
\[ \arctan y =\int_0^y (1-x^2+x^4-x^6+\cdots (-1)^nx^{2n})dx +R_n(y) \]
\[ \arctan y =y-\frac{y^3}{3}+\frac{y^5}{5}-\frac{y^7}{7}+\cdots+(-1)^n\frac{y^{2n+1}}{2n+1}+R_n(y) \]
As $R_n(y)\rightarrow 0$ ,
\[ \arctan y = y-\frac{y^3}{3}+\frac{y^5}{5}-\cdots \]
In this place, we put $y=1$ .
\[ \arctan 1 =\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \]
Therefore,
\[ \pi=4\left( \lim_{n\rightarrow \infty}\sum_{k=0}^n(-1)^k\frac{1}{2k+1} \right) \]
We got the series expression of $\pi$ .
