Tricky Integral Solution: How to Compute with Complex Analysis | a Real Number

[zpconn]

Homework Statement

How might one proceed to compute the integral depicted in the following image?

http://yfrog.com/jdintegralup

a is an arbitrary real number.

Homework Equations

I think the methods of complex analysis can be used.

The Attempt at a Solution

This integral, as far as I can tell, does not fall under any of the standard categories of integrals that are calculated using residues. Nevertheless I have a suspicion that residues can be used. Any method is of course fair game!

At this point I've just been fiddling with trying to find contours to integrate over, etc., without any luck. Experimentation suggests the answer should be π tanh(aπ/2).

 

[zpconn]

For anyone who has Basic Complex Analysis by Marsden and Hoffman, I've discovered this is problem 37 on Pg. 317. So it should be doable using residues since that's what that chapter covers. (Although the chapter also talks a bit about infinite sums, so it's conceivable that this might use those as well--I'm not sure how at this point.)

 

[iRaid]

integral -infinity infinity sin(a x) sinh(x) dx = (-infinity (cosh(x) sin(a x)-a sinh(x) cos(a x)))/(a^2+1)+constant

Hope it helped.

 

[zpconn]

Thanks for the response. Unfortunately I think you misread the integral: it is the quotient of sine and the hyperbolic sine, not the product.

 

[Count Iblis]

It is very easy to compute. You can replace sin(ax) by exp(iax), but before you can do that, first take the principal part. And then you add a small circle passing above the origin connecting the two parts of the integral.

Then consider the contour from minus R to plus R (that avoids the origin), then from R to R + pi i, from there to -R + pi, avoiding the point pi i on the imaginary axis and then back to -R. If you pass below the point pi i, then the contour integral will be zero. But the two half corcles you use to avoid the poles will make a net contribution. The upper part of the rectangle is related in a simple way to the lower part, thus allowing you to compute the integral.

 

[zpconn]

I question whether it's okay for me to bump this thread in this manner, but I just wanted to say thanks to Count Iblis for a correct and very simple method of computing this integral.

 

[Mark44]

I doubt very much that anyone would complain about bumping a thread to give thanks for help received.

 

[Count Iblis]

I question whether it's okay for me to bump this thread in this manner, but I just wanted to say thanks to Count Iblis for a correct and very simple method of computing this integral.

You are welcome