Two Indefinite Integrals Looking for some hints to evaluate the following integrals (with complex analysis or otherwise):
$$\int_0^\infty\frac{x^{p-1}}{x+1}\,dx,\;\;\;\; 0<p<1,$$
$$\int_{-\infty}^\infty e^{-s^2+isz}\,ds,\;\;\;\;z\in\mathbb{C}.$$
Thanks, I'm pretty stuck on both.
 A: For the first integral, you can set $t=\dfrac{x}{x+1}$ so that $\mathrm{d}x=\dfrac{\mathrm{d}t}{(1-t)^2}$, and you have

 $$\int_0^\infty \frac{x^{p-1}}{x+1}\,\mathrm{d}x=\int_0^1t^{p-1}(1-t)^{-p}\,\mathrm{d}t=\mathrm{B}(p,1-p)$$
 where $\mathrm{B}(a,b)$ denotes the Beta function.

For the second integral, consider completing the square, then recall that
$$\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x=\sqrt\pi$$
A: $$
s^2 - isz = \left(s^2 - isz - \frac{z^2} 4\right) + \frac{z^2} 4 = \left( s - \frac{iz} 2 \right)^2 + \frac{z^2} 4
$$
\begin{align}
& \int_{-\infty}^\infty e^{-s^2+isz}\,ds = e^{z^2/4}\int_{-\infty}^\infty e^{-(s - iz/2)^2} \, ds = e^{z^2/4} \int_{-\infty-iz}^{\infty - iz} e^{-t^2} \, dt \\[10pt]
= {} & e^{z^2/4} \lim_{M\to\infty} \int_{-M-iz}^{M - iz} e^{-t^2} \, dt \\[10pt]
= {} & e^{z^2/4} \lim_{M\to\infty} \left( \int_\gamma e^{-t^2} \, dt - \int_{-M}^M e^{-t^2}\,dt - \int_M^{M-iz} e^{-t^2} \, dt -\int_{-M-iz}^{-M} e^{-t^2} \, dt \right)
\end{align}
where $\gamma$ is a path along four straight lines: first from $-M$ to $M$, then from $M$ to $M-iz$, then from $M-iz$ to $-M-iz$, then from $-M-iz$ to $-M$.
We know that
$$
\int_\gamma e^{-t^2} \, dt = 0
$$
because it's the integral of an entire function along a closed path.
The value of
$$
\lim_{M\to\infty} \int_{-M}^M e^{-t^2}\,dt
$$
is well known.
If you can find the limits of the other two integrals, then you've got it.
A: First rewrite it as
$$ \displaystyle \int_{0}^{\infty}\dfrac{1}{{(x+1)x^{1-p}}}$$
If $x\in (0,1)$ then $x^{2-p}+x^{1-p}>x^{1-p}$. Therefore 
$$ \displaystyle \int_{0}^{1}\dfrac{1}{{(x+1)x^{1-p}}}dx<\int_{0}^{1}\dfrac{1}{{x^{1-p}}}dx$$
Since the larger one converges by $p$ test the smaller one converges by comparison test.
If $x\in (1,\infty)$ then $x^{2-p}+x^{1-p}>x^{2-p}$. Therefore 
$$ \displaystyle \int_{1}^{\infty}\dfrac{1}{{(x+1)x^{2-p}}}dx<\int_{1}^{\infty}\dfrac{1}{{x^{2-p}}}dx$$
Since the larger one converges by $p$ test the smaller one converges by comparison test.
