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In a recent lecture a professor told a story about the integral below. Lobachevsky calculated this integral at first time incorrectly. Following the publication of the integral, Ostrogradsky sent a letter with correct answer to Lobachevsky.

What is the right answer?

$$I(a)=\int_{0}^{\infty}\frac{e^x-e^{-x}}{e^{2x}+e^{-2x}+2cos(2a)}xdx;0\leq a \leq \pi$$

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Do you know the answer? –  Fabian Apr 5 '11 at 7:15
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What was the incorrect result? –  Fabian Apr 5 '11 at 8:35
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1 Answer

up vote 7 down vote accepted

We can restrict the values of $a$ to be between $0$ and $\pi/2$ as $\cos(2a)= \cos[2(\pi-a)]$. With this $$I(a) = \frac{\pi a}{4 \sin(a)} \qquad 0\leq a \leq \frac{\pi}{2}.$$

The calculation can be done along the following line:

  • As the integrand is symmetric the integration region can be extended to the full real line $$I(a) = \frac{1}{2} \int_{-\infty}^{\infty}dx\, \frac{(e^x-e^{-x})x}{e^{2x}+e^{-2x}+2\cos(2a)}.$$

  • The substitution $z= e^{x}$ brings the integral onto the form $$I(a) = \frac{1}{2} \int_0^\infty dz\,\frac{(z^2-1) \log z}{z^4 + 2 z^2 \cos(2a) + 1}.$$

  • A standard trick can be employed to bring it on the form $$I(a) = \frac{1}{4} \sum_{z_n} \,\text{Res}_{z=z_n} \frac{(z^2-1) \log^2 z}{z^4 + 2 z^2 \cos(2a) + 1}$$ where $z_n$ are the zeros of $z^4 + 2 z^2 \cos(2a) + 1$ and the branch cut of $\log$ is along the negative real line.

  • The 4 zeros of $z^4 + 2 z^2 \cos(2a) + 1$ are given by $\bar z=\pm i e^{\pm i a}$. The residues assume the form $$\text{Res}_{z=\bar z} \frac{(z^2-1) \log^2 z}{z^4 + 2 z^2 \cos(2a) + 1} = \frac{(\bar z^2 - 1)\log^2 \bar z}{4 \bar z^3 +4 \bar z \cos 2a}.$$

  • Putting everything together, we obtain the result quoted above (after some tedious but straightforward calculation).

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Incorrect answer was $I(a) = \frac{\pi a}{4 \sin(a)} \qquad 0\leq a \leq \pi$. –  Martin Gales Apr 6 '11 at 7:02
    
@Martin Gales: do you know an easier way to calculate the final answer....? –  Fabian Apr 6 '11 at 15:02
    
The professor gave your question as a challenge. Ostrogradsky solved this integral without complex analysis in a very short way he said. I have no a clue at this moment. –  Martin Gales Apr 8 '11 at 5:46
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