My Math Forum Integral - Real in C

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December 2nd, 2010, 01:08 PM   #11
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Re: Integral - Real in C

Spoiler! I asked about the problem. I was referred to a table of formulas that came with the exam. Strange! However, we did an exercise earlier in the course.

$\displaystyle \int_0^\infty \frac{1}{(1+x)x^a} dx$ $0 < a < 1$

The solution given in the lecture is as follows.

A closed path, C, is used to calculate the residue in -1. See the figure in the attached file. $\displaystyle \int_C$ = $2\pi i Res[f(x);x=-1]$ = $2\pi i (\frac{1}{x^a})_{x=-1}$ = $2\pi i e^{-i\pi a}$

Now I and II are examined. With the help of. $x^a$ = $e^{a \log x}$ = $e^{ln |x| + i \arg x}$ = $|x|^a e^{i \arg x * a}$ = $|x|^a e^{2 \pi i a}$
$I= \int_r^R \frac{1}{(1+x)x^a} dx$ =[color=#BF0000] ??? not calculated I guess it's solved further down.[/color]

$II= \int_R^r \frac{1}{(1+x)x^a} dx$ = $I(-e^{-2\pi i a})$ [color=#BF0000] ??? Is this the difference between I and II?[/color]

$\left|\oint_{C_R} \right| \leq 2 \pi R \frac{1}{(R-1) R^a} \rightarrow 0$ if $R \rightarrow \infty$

$\left|\oint_{C_r} \right| \leq 2 \pi r \frac{1}{(r-1) r^a}$ = $r^{1-a} \frac{2\pi}{1-r} \rightarrow 0$ if $r \rightarrow 0$ [color=#BF0000]??? I don't understand this[/color]

This sums up like this.

$I + I \left( -e^{-2\pi i a}\right)$ = $2\pi i e^{-\pi i a}$

$I= \frac{2 \pi i e^{-i \pi a}}{1 - e^{-2 \pi i a}}$ = $\frac{2 \pi i }{e^{i\pi a} - e^{- \pi i a}}$ = $\frac{\pi}{\sin (\pi a)}$
Attached Images
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 December 2nd, 2010, 01:10 PM #12 Newbie   Joined: Nov 2010 From: Sweden Posts: 18 Thanks: 0 Re: Integral - Real in C I didn't post my question... Would it be hard to calculate the integral if a was$0 < a < \infty$
December 3rd, 2010, 12:10 PM   #13
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Re: Integral - Real in C

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 December 4th, 2010, 06:39 AM #14 Newbie   Joined: Nov 2010 From: Sweden Posts: 18 Thanks: 0 Re: Integral - Real in C Thank a million! I'll have to study you're pdf a bit more to fully grasp it.

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