How Does Contour Integration of Complex Functions Help Compute Integrals?

[Math Monster]

Homework Statement

Show that the formula:
∫_0^∞ (s^a-1)/(s-1) ds = -pi cot (a pi)
may be calculated by considering an analytical branch of function:
f(z) = z^(a-1) / (z-1)
and integrating along a contour consisting of:
a) a circle radius R, centred at (0,0)
b) with a branch cut running from (0,0) to R above and below the x-axis
c) circular contours around singularity (0,0) and (1,0)
when R→∞

I have attempted to substitute s=Re^i alpha z, and I believe the integral around (a) → 0 as R → infinity but i can't really make sense of it all!

Please help and thanks in advance!

 

[jackmell]

Homework Statement

Show that the formula:
∫_0^∞ (s^a-1)/(s-1) ds = -pi cot (a pi)
may be calculated by considering an analytical branch of function:
f(z) = z^(a-1) / (z-1)
and integrating along a contour consisting of:
a) a circle radius R, centred at (0,0)
b) with a branch cut running from (0,0) to R above and below the x-axis
c) circular contours around singularity (0,0) and (1,0)
when R→∞

I have attempted to substitute s=Re^i alpha z, and I believe the integral around (a) → 0 as R → infinity but i can't really make sense of it all!

Please help and thanks in advance!

Is it:

$$\text{P.V}\int_0^{\infty}\frac{s^{a-1}}{s-1}ds,\quad 0<a<1$$

If so, then just work a simple one first: Let $a=1/2$ and consider

$$\text{P.V}\int_0^{\infty} \frac{1}{\sqrt{s}(s-1)}ds$$

That's 8 integrals right? Can you set up some of them?

 

[Math Monster]

NB I tried to write this as Latex but not sure how to transfer it! If you can give me a hint I will try to repost so that it is easier to read!

I think so...

This has singularies at s=0 and s=1, same as my example.

Start with $s = e^{iz}$ and so $ds = ize^{iz} dz$
I've left it in terms of s for ease of typing, but realize that this transformation should be made (if I'm right that is!).

If i break my contour up into
$C_{R2} : \\*
\int^{-R}_{R} \frac{1}{\sqrt{s}(s-1)} ds $

$C_{1}: \\*
\int^{1+ε}_{R} \frac{1}{\sqrt{s}(s-1)} ds \\*
NB: y has no relevance, i.e. z=x, and so we integrate over x $

$C_{r_1}: \\*
\int_{1+ε}^{1-e} \frac{1}{\sqrt{s}(s-1)} ds \\*
Thinking maybe this should be integrated 0:pi?

$C_{3}: \\*
\int_{1-ε}^{0+ε} \frac{1}{\sqrt{s}(s-1)} ds \\*
NB: y has no relevance, i.e. z=x, and so we integrate over x $

$C_{r_2}: \\*
\int_{0}^{2 pi} \frac{1}{\sqrt{s}(s-1)} ds \\*
Not sure about this one? I think maybe we do over dz? And the circle should have radius ε?

$C_{4}: \\*
\int^{1-ε}_{0+ε} \frac{1}{\sqrt{s}(s-1)} ds \\*
NB: y has no relevance, i.e. z=x, and so we integrate over x $

$C_{r_{-1}}: \\*
\int^{1+ε}_{1-e} \frac{1}{\sqrt{s}(s-1)} ds \\*
Following from C_{4} I'm thinking maybe this should also be integrated 0:pi?

$C_{1}: \\*
\int_{1+ε}^{R} \frac{1}{\sqrt{s}(s-1)} ds \\*
NB: y has no relevance, i.e. z=x, and so we integrate over x $

Then the integral of the contour is the sum of these 8.


Am I anywhere near the right lines?

Also, just wanted to confirm that yes, 0<a<1.

Many thanks

 

[Whovian]

Use $and$, or [tex] if you want it to be in the centre of the screen, on its own line. Using [itex], I suggest you use \displaystyle.

 

[jackmell]

You need to enclose math code in [itex]\text{[tex]}[/itex] and [itex]\text{[/tex]}[/itex]. Tell you what, If I start with the leg from 0 to 1 above the real axis, go to infinity, around the circle $z=Re^{it}$, and then over the leg below the real axis back to the origin, then around it via $z=\rho e^{it}$, I get by the Residue Theorem:

$$\text{P.V.}\int_0^1 f(z)dz-\pi i r_a+\text{P.V} \int_1^{\infty} f(z)dz+\mathop\int\limits_R f(z)dz+\text{P.V} \int_{\infty}^1 (-f(z))dz-\pi i r_b+\text{P.V}\int_1^0 (-f(z))dz+\mathop\int\limits_{\rho} f(z)dz=0$$

Note the contours over the indentaitons around the pole at 1 is just -pi i times it's residue. Just for now, say the contour over the big and little circle is zero. Justify that for the purist later. Also, when we go around the branch point and onto the leg below the real axis, the value of the root changes by $e^{2\pi i/2}$. That gives the negatives along those contours. If it was $\sqrt[3]{z}$, and we go around once, it would change by $e^{2\pi i/3}$ and so forth. Keep that in mind when you consider $a=2/3$. Combine all those principal values into one cus' it's the same thing and then write:

$$2\;\text{P.V.}\int_0^{\infty} f(z)dz-\pi i(r_a+r_b)=0$$

So
$$\text{Res} (\frac{1}{\sqrt{z}(z-1)},1)=\pm 1$$

corresponding to the two roots of 1 right? Get that one straight, then adapt it to $a=2/3$. That one is not zero.