## Monday Math 38

Find .

The methods I’ve shown before for definite integrals (differentiating under the integral sign, converting to a double integral and reversing order, expanding in infinite series) will not be of much use on this one. However, we can use complex analysis and contour integration to find this.

First, due to symmetry, we see . Further, symmetry also tells us , as the integrand is odd. Thus:
.

So we will compute the integral of  over an appropriate contour in the complex plane. As , the function has poles at ±i. Expanding in partial fractions, , and so we see the residues of  at z=i and z=-i are  and , respectively.

Let us first choose the (counter-clockwise) contour bounding a semicircle of radius R>1 in the upper half plane; that is, the contour consists of the real axis from –R to R, and then the semicircle of radius R in the upper half plane. We can parametrize the semicircle as , 0≤t≤1.

The contour encloses only the pole at z=i, so therefore the integral of our function over our contour is given by the residue theorem as:


Breaking up our contour into its two segments, though, we see:
, where C1 is our semicircle.
Thus 

Now, we will need to use the following theorem of complex analysis:
Suppose f(z) is continuous on the contour C, and there exists a real constant M such that  on C. Then , where L is the length of the contour C.

So, examining |f(z)| on C1, we see:
.
For a>0, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1, and thus
.
Since the length of our semicircle is πR, our theorem tells us
. Now, as R goes to infinity,  goes to zero. Thus, .

Applying this limit to our equation for the real axis portion,

So, for a>0,
.

Note that if a<0, our bound on the semicircle does not work. So now, let us perform the integral over the contour bounding a semicircle in the lower half-plane; it consists of the real axis from R to -R, and then the semicircle of radius R in the lower half plane, which we can parametrize as , 0≤t≤1.

This contour encloses the pole at z=-i only, so via the residue theorem,


Breaking up our contour into its two segments, though, we see:
, where C2 is our semicircle.
Thus .

As to |f(z)| on C2, we see:
.
Here, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1 when a<0, and thus
.
The length of our semicircle is still πR, and so applying the theorem and taking the limit as R goes to infinity, we get
.
Therefore , and so  for a<0.

Combining these two, we see:
.

(Note that a=0 gives the correct equation ).