Estimating the integral

Question

I'm trying to find an estimate for the modulus of the integral given below.
|$\int_0^r\frac{w^{z-1}}{w^2+1}dw$|, where $r\in[0,1]$ and $z\in\mathbb{C}$ such that $Re(z)>0$.
Any hint will be greatly appreciated. Thanks!

Answer 1

Write $1/(w^2+1) =\sum_{n=0}^{\infty} (-1)^n w^{2n} $, multiply, and integrate term by term. The result will be a series with alternating lower and upper bounds.

(added later)

Since $\dfrac{1-x^n}{1-x} =\sum_{k=0}^{n-1} x^k $, putting $-x$ for $x$ gives $\dfrac{1-(-1)^nx^n}{1+x} =\sum_{k=0}^{n-1} (-1)^kx^k $, so $\dfrac{1-(-1)^nx^{2n}}{1+x^2} =\sum_{k=0}^{n-1} (-1)^kx^{2k} $ so $\dfrac{1}{1+x^2} =\dfrac{(-1)^nx^{2n}}{1+x^2}+\sum_{k=0}^{n-1} (-1)^kx^{2k} $.

Now we can integrate.

$\begin{array}\\ \int_0^r\dfrac{w^{z-1}}{w^2+1}dw &=\int_0^rw^{z-1}\left(\sum_{k=0}^{n-1} (-1)^kw^{2k}+\dfrac{(-1)^nw^{2n}}{1+w^2}\right)dw\\ &=\sum_{k=0}^{n-1} (-1)^k\int_0^rw^{z-1}w^{2k}dw+\int_0^rw^{z-1}\dfrac{(-1)^nw^{2n}}{1+w^2}dw\\ &=\sum_{k=0}^{n-1} (-1)^k\int_0^rw^{2k+z-1}dw+\int_0^r\dfrac{(-1)^nw^{2n+z-1}}{1+w^2}dw\\ &=\sum_{k=0}^{n-1} (-1)^k\dfrac{w^{2k+z}}{2k+z}|_0^r+(-1)^nE_{2n+z-1}(r) \qquad E_{v}(r)=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\ &=\sum_{k=0}^{n-1} (-1)^k\dfrac{r^{2k+z}}{2k+z}+(-1)^nE_{2n+z-1}(r)\\ \end{array} $

Here are some easy bounds for $E_v(r)$.

$\begin{array}\\ E_{v}(r) &=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\ &<\int_0^rw^{v}\\ &=\dfrac{r^{v+1}}{v+1}\\ E_{v}(r) &=\int_0^r\dfrac{w^{v}}{1+w^2}dw\\ &>\int_0^r\dfrac{w^{v}}{1+r^2}dw\\ &=\dfrac{r^{v+1}}{(v+1)(1+r^2)}\\ \end{array} $