1. ## Complex Analysis

Hi everyone, I have a big test tomorrow and I'm still hung up on integrating over branch cuts.

So I was wondering how to show

$\displaystyle \int_0^{\infty} \frac{t^{\alpha-1}}{t+1} dt = \frac{\pi}{\sin(\pi\alpha)}$
where $\displaystyle 0 < \alpha <1$

2. Originally Posted by chiph588@
Hi everyone, I have a big test tomorrow and I'm still hung up on integrating over branch cuts.

So I was wondering how to show

$\displaystyle \int_0^{\infty} \frac{t^{\alpha-1}}{t+1} dt = \frac{\pi}{\sin(\pi\alpha)}$
where $\displaystyle 0 < \alpha <1$
Consider the integral,
$\displaystyle \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx \text{ where }0 < a < 1$
If we let $\displaystyle t = \log x$ then we get,
$\displaystyle \int \limits_0^{\infty} \frac{t^{a-1}}{t+1} dt \text{ where }0 < a < 1$
Therefore, to evaluate your integral is is sufficient to evaluate the first (more convient) integral.

Fix $\displaystyle R>0$ (and "large" i.e. large enough for all the inequalities to work out)
let $\displaystyle \Gamma$ be the rectangular contour with vertices $\displaystyle \pm R, \pm R + 2\pi i$.

Now define the function $\displaystyle f(z) = \frac{e^{az}}{e^z + 1}$ the function has a pole at $\displaystyle z=\pi i$ with $\displaystyle \text{res}(f,\pi i) = - e^{\pi a i}$.

Integrating by the residue theorem we get that,
$\displaystyle \int \limits_{-R}^R \frac{e^{ax}}{e^x + 1} dx + \int_0^{2\pi} \frac{e^{a(R+it)}}{e^{R+it} + 1} dt - \int \limits_{-R}^R \frac{e^{a(x+2\pi i)}}{e^{x+2\pi i} +1} dx - \int_0^{2\pi} \frac{e^{a(-R +it)}}{e^{-R+it} + 1} dt = -2\pi i e^{\pi a i }$

We will now show that the 2nd and 4th integrals (over the vertical sides) go to zero as $\displaystyle R\to \infty$.
Look at the function inside the 2nd integral,
$\displaystyle \left| \frac{e^{a(R+it)}}{e^{R+it} + 1} \right| \leq \frac{|e^{a(R+it)}|}{||e^{R+it}| - |1||} = \frac{e^{aR}}{e^R - 1} \leq \frac{e^{aR}}{e^R - \frac{1}{2}e^R} = 2e^{(a-1)R}$
Look at the function inside the 4th integral,
$\displaystyle \left| \frac{e^{a(-R+it)}}{e^{-R+it} + 1} \right| \leq \frac{|e^{a(-R+it)}|}{||e^{-R+it}| - |1||} = \frac{e^{-aR}}{1 - e^{-R}} \leq \frac{e^{-aR}}{1 - \frac{1}{2}} = 2e^{-aR}$
Therefore the 2nd integral satisfies,
$\displaystyle \left| \int_0^{2\pi} \frac{e^{a(R+it)}}{e^{R+it} + 1} dt \right| \leq 4\pi e^{(a-1)R} \to 0 \text{ since }a-1 < 0$
Therefore the 4th integral satisfies,
$\displaystyle \left| \int_0^{2\pi}\frac{e^{a(-R+it)}}{e^{-R+it} + 1} \right| \leq 4\pi e^{-aR} \to 0 \text{ since } -a < 0$
Look at the 3rd integral,
$\displaystyle \int \limits_{-R}^R \frac{e^{a(x+2\pi i)}}{e^{x+2\pi i}+1} dx = e^{2\pi a i}\int \limits_{-R}^R \frac{e^{ax}}{e^x +1} dx$
Taking the limits in the big equation with four integral above we get,
$\displaystyle (1 - e^{2\pi a i} ) \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx = - 2\pi i e^{\pi a i}$
Finally,
$\displaystyle \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx = \frac{2\pi i e^{\pi a i}}{e^{2\pi a i} - 1} = \frac{2\pi i}{e^{\pi a i} - e^{-\pi a i}} = \frac{\pi}{\sin (\pi a)}$
----

At this point we can derive the Euler Reflection Formula.
Remember by the first comment we have shown that,
$\displaystyle \int_0^{\infty} \frac{t^{a-1}}{t+1} dt = \frac{\pi}{\sin (\pi a)}$
Let $\displaystyle \mu = \tfrac{t}{t+1}$ to get,
$\displaystyle \int_0^{\infty} \mu ^{a-1}(1-\mu)^{-a} d\mu = \bold{B}(a,1-a) = \Gamma (a) \Gamma (1-a) = \frac{\pi}{\sin (\pi a)} \text{ for }0 < a < 1$

3. Hello,

$\displaystyle \int_0^{\infty} \frac{t^{-\alpha}}{1+t} ~ dt=\int_0^{\infty} \frac{1}{t^\alpha (1+t)} ~ dt$

Lemma : (actually, I only know the French naming of this...)

$\displaystyle 2i \pi \sum_{w \in \mathbb{C} \backslash \mathbb{R}_+} \text{Res}_w \left(\frac{R(z)}{z^\alpha}\right)=\left(1-e^{-2i \pi \alpha}\right) \int_0^\infty \frac{R(x)}{x^\alpha} ~ dx$
where R is a rational fraction without poles on the real axis $\displaystyle x \geq 0$ and $\displaystyle R(x) \stackrel{|x|\to \infty}{\sim} \frac{1}{x^n},~ n \geq 1$. $\displaystyle 0<\alpha<1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This can be proved by considering the contour integral :
$\displaystyle \oint \frac{R(z)}{z^\alpha} ~ dz$ over this contour :

as $\displaystyle e \to 0,~ h \to 0,~ R \to \infty$
Let $\displaystyle f(z)=\frac{R(z)}{z^\alpha}=\frac{R(z)}{e^{\alpha \log(z)}}$

$\displaystyle \bullet \quad$ When t goes to infinity, $\displaystyle |f(t)| \leq \frac{c}{|z|^{1-\alpha}}$. Then $\displaystyle \left|\int_\Gamma f(z) ~ dz\right| \leq \frac{c}{R^{1+\alpha}} \times R=\frac{c}{R^\alpha} \stackrel{R \to \infty}{\longrightarrow} 0$

$\displaystyle \bullet \quad |R(z)| \leq \frac{c}{|z|^\alpha}$. When z is near from 0, we thus have $\displaystyle \left|\int_\gamma f(z) ~ dz \right| \leq c' \epsilon^{1-\alpha} \stackrel{\epsilon \to 0}{\longrightarrow} 0$

$\displaystyle \bullet \quad \int_{\Gamma_1} f(z) ~ dz \to \int_0^{\infty} \frac{R(x)}{x^\alpha} ~ dx=I$

$\displaystyle \bullet \quad \int_{\Gamma_2} f(z) ~ dz \to \int_0^{\infty} \frac{R(x)e^{-2i \pi \alpha}}{x^\alpha} ~ dx=e^{-2i \pi \alpha} I$
The $\displaystyle 2i \pi \alpha$ comes from the fact that :
$\displaystyle z \to x,~ \Im z>0,~ \log z \to \ln(x) \implies z^\alpha \to x^\alpha$
But $\displaystyle z \to x, ~ \Im z<0,~ \log z \to \ln(x)+2i \pi \implies z^\alpha \to e^{2i \pi \alpha} x^\alpha$

Hence, by the residue theorem, we have :
$\displaystyle I-e^{-2i \pi \alpha} I=\int_{\Gamma \cup \gamma \cup \Gamma_1 \cup \Gamma_2} f(z) ~ dz=2i \pi \sum_{w \in \mathbb{C} \backslash \mathbb{R}_+} \text{Res}_w \left(\frac{R(z)}{z^\alpha}\right)$
$\displaystyle \blacksquare$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There is only one pole over $\displaystyle \mathbb{C} \backslash \mathbb{R}_+$ : $\displaystyle t=-1$
$\displaystyle \text{Res}_{-1} ~ \frac{1}{t^\alpha(1+t)}=\frac{1}{(-1)^\alpha}$

We have $\displaystyle \log(-1)=i \pi$
Hence $\displaystyle (-1)^\alpha=e^{i \pi \alpha}$

By the above lemma, we get :
$\displaystyle 2i \pi \cdot e^{-i \pi \alpha}=(1-e^{-2 i \pi \alpha}) I$

Hence :
$\displaystyle \frac{I}{\pi}=\frac{2i}{e^{i \pi \alpha}-e^{-i \pi \alpha}}=\frac{1}{\sin(\pi \alpha)}$

$\displaystyle \boxed{I=\frac{\pi}{\sin(\pi \alpha)}}$

Okay, there may be mistakes and I'm sure there are unclear things, but one can just remember the formula of the lemma. Sorry for the very long message

4. If you're interested, I found completely by chance a wikipedia version of what I did : Mellin transform - Wikipedia, the free encyclopedia