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In the following paper (page 10, section 7), the authors state that $\displaystyle\int_{-\infty}^{\infty} \frac{1}{(b^{2}+x^{2})\cosh ax} \ dx $ can be evaluated by "closing the real axis with a semi-circle centered at the origin located in the upper half-plane. An elementary estimate shows that the integral over the circular boundary vanishes as the radius goes to infinity."

The integrals in Gradshteyn and Ryzhik. Part 21: Hyperbolic functions

Does the integral over the circular boundary really evaluate to zero when you take the limit? I don't think it does. Wouldn't a rectangle be the appropriate contour for this problem?

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The top half of the circle tends to zero. The real part does not. A semicircle is the simplest contour to use for this integral. –  Argon Apr 6 '12 at 19:54

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You're right, the argument in the paper isn't rigorous because one would have to consider what happens when $|\cosh ax|\lt1$. Continously taking either a semicircle or a rectangle to infinity would cross the poles. You're also right that showing how to avoid this complication is easier for a rectangle than for a semicircle. Since $\cosh (x+\mathrm iy)=\cosh x\cos y+\mathrm i\sinh x\sin y$, rectangles at arbitrary distances from the origin can be chosen such that $|\cosh ax|\ge1$, and then the factor $b^2+x^2$ in the denominator ensures convergence.

Of course, since the integrals along the rectangular contours converge, the integrals along the semicircular contours also converge, since they're equal if they enclose the same poles; but this would be more difficult to show.

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Say $b=1$ and $a = \pi $. My idea would be to use a rectangle with vertices at $N, N+iN,-N+iN,$ and $-N$ where $N$ is an integer greater than $1$ so that I'm not going through a pole. Then showing the integral tends to zero around the three other sides is an easy application of the ML inequality. But if that works, then I guess I could instead use a circle with a radius of $N$. –  Random Variable Apr 6 '12 at 20:48
    
For some reason it didn't occur to me that if a rectangular contour would work, then a semi-circular contour would work, too. –  Random Variable Apr 6 '12 at 20:59
    
@RandomVariable: Yes, but if you did it directly using a semi-circular contour, you'd need a slightly more complicated argument. –  joriki Apr 6 '12 at 21:14
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@RandomVariable: I think that was a misunderstanding. I had argued from the beginning that the integrals along the semi-circles must converge because the integrals along the rectangles do. I was merely saying that the latter is easier to show directly, whereas the convergence of the integrals along the semi-circles is easy to infer from the convergence of the integrals along the rectangles, but is more difficult to rigorously prove directly. –  joriki Apr 6 '12 at 22:22
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@RandomVariable: Yes, the modulus of $\cosh az$ increases exponentially with the real part of $z$. And yes, the ML inequality is the basis for all this; but to easily apply it we need $|\cosh ax|\ge1$, which we can ensure easily at the top of the rectangle but not as easily along the semi-circle. –  joriki Apr 6 '12 at 22:24

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