Catalan constant preliminaries

Question

I recently ran across this representation of :

$$ \Bigg[ \frac{1}{1+x^{2}}=\frac{1}{x^{2}(1+\frac{1}{x^{2}})=\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...)}\Bigg]$$

I believe the nonending term in the denominator comes from long division but I do not understand how this is derived. Any help is welcomed.

I also saw a different form of the $\arctan(x)$ integral and I don't understand where it comes from. The $\arctan(x)$ integral:

$$\int_0^x \frac{dy}{(1+y^{2})}$$ can be written as: $$ \arctan(x)= \int_0^x (1-y^{2}+y^{4}-y^{6}+y^{8}-...)dy$$ Evidently the integrand as written comes from a long division process derived from the original integrand but I don't see how. Again, any help would be welcomed.

Answer 1

Note that $1-z+z^2-z^3+z^4-z^5+\cdots$ is a geometric series with the common ratio of $-z$. Also, $1+r+r^2+\cdots+r^n = \frac{1-r^{n+1}}{1-r}$. Consequently, if $|z|<1$ then $$1-z+z^2-z^3+z^4-z^5+\cdots = \lim_{n\to\infty}\frac{1-(-z)^n}{1+z}= \frac{1}{1+z}.$$ In your examples, $z = 1/x^2$ or $y^2$.

Answer 2

Take $y=\tan u $so $$dy=(1+\tan ^2 u) du $$ put into $\int \frac{dy}{(1+y^{2})}$ an you will have $$\int \frac{dy}{(1+y^{2})}=\int \frac{(1+\tan ^2 u) du}{1+\tan ^2 u }$$simplify $$\int \frac{(1+\tan ^2 u) du}{1+\tan ^2 u }=\int 1du =u$$ now $$y=\tan u \to u=\arctan y\\\int_0^x\frac{dy}{(1+y^{2})}=\arctan y|^x_0=\arctan x $$

Answer 3

The first one is malformed.

$$\Bigg[ \frac{1}{1+x^{2}} =\frac{1}{x^{2}(1+\frac{1}{x^{2}}) =\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...)}\Bigg] $$ should be $$ \frac{1}{1+x^{2}} =\frac{1}{x^{2}(1+\frac{1}{x^{2}})} =\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...) $$