2

The problem statement is:

$$\int_0^{\infty}\frac{x^α}{x^3+1}dx$$ for α in the range −1<α<2.

$$\huge \frac{2\pi i}{1-e^{\frac{i2\pi (\alpha+1)}{3}}} \frac {e^{\frac{i \pi \alpha}{3}}} { 3e^{\frac{2\pi i}{3}}}$$

$\alpha$ is some constant between -1 and 2.

Thanks,

Martin Sleziak
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user282725
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  • That is $-\dfrac{\frac\pi3}{\sin\left(\frac\pi3a\right)}e^{-\frac{2\pi i}3} =\dfrac{\frac\pi3}{\sin\left(\frac\pi3a\right)}e^{\frac{\pi i}3} =\dfrac{\frac\pi3}{\sin\left(\frac\pi3a\right)} \left(\frac12+\frac{\sqrt3}2i\right)$, which is not real. – robjohn Oct 22 '15 at 09:59
  • Is $\alpha$ a real constant or can it be complex? – Horus Oct 22 '15 at 10:05
  • This does not seem to be [tag:integration] or [tag:residue-calculus]. – robjohn Oct 22 '15 at 10:07
  • I finally found it. f-bomb. Thanks @robjohn. It is residue-calculus. A good all-nighter's worth of work and probably not wise to spin my wheels that long on a problem before asking a question on MSE. Thanks so much :-) – user282725 Oct 22 '15 at 10:09
  • Meaning I spent all night believing that it must be real, and my computations must be correct, since Wolfram Alpha didn't give an answer, when I went to check with it and instead it gives "standard computation time exceeded", and since you confirmed that it is not a real number, I looked back at my work and found the issue. Hard lesson learned. Thanks @RobJohn. – user282725 Oct 22 '15 at 10:14
  • Maybe the problem that led up to this computation was a residue-calculus problem, but if you don't include more context, this question does not, by itself, involve residue-calculus. – robjohn Oct 22 '15 at 10:15
  • Yes, understood - sorry... – user282725 Oct 22 '15 at 10:16
  • Hi @robjohn, I just edited my question to update the number. Is this now real? I added 1 to $\alpha$ ... thanks, – user282725 Oct 22 '15 at 10:33
  • And if it is real, can you offer a hint on how to best show it? Thanks @robjohn – user282725 Oct 22 '15 at 10:44
  • Now it looks like $\dfrac{\frac\pi3}{\sin\left(\frac\pi3(a+1)\right)}$ – robjohn Oct 22 '15 at 10:45
  • Awesome! It's real!! I just posted the integral too. How do you think I can approach showing the number is real? Just expand out all the Euler formulas, or is there a more clever way? Thanks @robjohn – user282725 Oct 22 '15 at 10:48

2 Answers2

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Multiply by $\frac{e^{\frac{-i\pi(a+1)}3}}{e^{\frac{-i\pi(a+1)}3}}$: $$ \begin{align} \frac{2\pi i}{1-e^{\frac{i2\pi(a+1)}3}}\frac{e^{\frac{i\pi a}3}} {3e^{\frac{2\pi i}3}} &=\frac{2\pi i}{e^{-\frac{i\pi(a+1)}3}-e^{\frac{i\pi(a+1)}3}}\frac{e^{-\frac{i\pi}3}} {3e^{\frac{2\pi i}3}}\\[6pt] &=\frac{\frac\pi3}{-\sin\left(\frac\pi3(a+1)\right)}e^{-i\pi}\\[6pt] &=\frac{\frac\pi3}{\sin\left(\frac\pi3(a+1)\right)} \end{align} $$ since $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$.

robjohn
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  • This has come up so often in past problems (finalizing the answer to include a sine, cosine, sinh, or cosh), yet still so hard to see. Thanks so much for the cool technique, @RobJohn. Have a great night! :-) – user282725 Oct 22 '15 at 11:02
  • Take a look at this answer, which generalizes your integral. – robjohn Oct 22 '15 at 11:06
  • Ok awesome -- thanks so much for the link. I'll be sure to read it in more detail after some sleep. Thanks and good night @robjohn! :-) – user282725 Oct 22 '15 at 11:21
0

HINT:

Assuming $x\in\mathbb{R}$:

$$\frac{2\pi i}{1-e^{\frac{i2\pi x}{3}}} \frac {e^{\frac{i \pi x}{3}}} { 3e^{\frac{2\pi i}{3}}}=$$ $$\frac{\left(2\pi i\right)\cdot\left(e^{\frac{x\pi i}{3}}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i\left(\cos\left(\frac{\pi x}{3}\right)+\sin\left(\frac{\pi x}{3}\right)i\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{3(i+\sqrt{3})e^{\frac{i\pi x}{3}}\sin\left(\frac{\pi x}{3}\right)}$$

Jan Eerland
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