]> 27.2 Some Special Tricks

## 27.2 Some Special Tricks

There are functions that are not integrable in general but for which integrals between certain specific endpoints can be evaluated.

These are often integrals that can be rewritten, either by adding a known integral, or by using symmetry, or by some other trick, as an integral over a closed path in the complex plane.

Such integrals can be evaluated by use of the Residue Theorem, which states that the integral of a function $f ( z )$ counterclockwise around a simple closed path $C$ is $2 π i$ times the sum of the residues of $f$ within $C$ .

The residue of a function with an isolated singularity is the coefficient of its minus first power at the singular point.

Thus for example, $2 z$ has residue 2 at $z = 0$ . $1 sin ⁡ z$ has residue 1 at even multiples of $π$ , and residue -1 at the odd multiples of $π$ .

We know that the function $1 1 + x 2$ is the derivative of the arctangent, so we can actually integrate it over any range. In particular since we have $tan ⁡ ( π 2 ) = + ∞$ and $tan ⁡ ( − π 2 ) = − ∞$ , its integral from $− ∞$ to $+ ∞$ is $π$ .

Here is another way to deduce this fact. We can use partial fractions to write

$1 1 + x 2 = 1 ( x + i ) ( x − i ) = 1 2 i ( x − i ) − 1 2 i ( x + i )$

Thus this function has a singularity at $i$ in the upper half plane and a singularity at $− i$ in the lower half plane, with residues $1 2 i$ and $− 1 2 i$ respectively. If we make a path $C$ that goes up the real axis from $− R$ to $R$ then around a semicircle in the upper half plane from $R$ back to $− R$ , for $R > 1$ this will enclose the singularity at $i$ .

The value of the integral will therefore be $2 π i 2 i$ or $π$ , by the residue theorem.

The integrand will behave like $R − 2$ on the large semicircle, and since the length of that semicircle is only $π R$ , the integral around it will go to zero like $1 R$ does as $R$ increases.
This tells us that the value of the integral from $− R$ to $R$ on the real line will go to $π$ as $R$ increases.

This gives us an integral we know.

However the same technique applies to much more complicated integrands and allows us to do lots of integrals again from $− R$ to $R$ as $R$ approaches infinity.

We give two examples.

One is the so called Fourier transform of $1 1 + x 2$

$∫ − R R e i k x 1 + x 2 d x + ∫ C e i k z 1 + z 2 d z = 2 π i Re ⁡ s ( e i k z 1 + z 2 ) a t ( z = i ) = π e − k$

where $C$ is the semicircle of radius $R$ in the upper half plane, and again the integral on $C$ goes to zero as $R$ increases.

Now we use this method to sum a series. The function $cot ⁡ x$ is singular at $x = 0$ and is periodic with period $π$ so it is singular at every multiple of $π$ . Its residue at each of these singularities is 1.

Moreover, if you wander far off the real line, it quickly approaches either $i$ or $− i$ , since it is

$i e i x − y + e − i x + y e i x − y − e − i x + y$

and the second terms in the numerator and denominator will dominate in the upper half plane making the integrand approach $− i$ and the first terms will dominate in the lower half plane so that it approaches $i$ as $| y |$ increases there.

This implies that an integral of $cot ⁡ z z 2$ over a large circle of radius $R$ (with $R$ a half integral multiple of $π$ to avoid trouble near the real line) will go to zero as $R$ goes to infinity just as in the previous case.

This further implies that the sum of the residues of this function must go to zero inside this circle. But for each positive or negative integer $j$ the residue of this function at $j π$ is $( j π ) − 2$ .

The residue of this integrand at $z = 0$ can be computed as half the second derivative of $z cot ⁡ z$ at $z = 0$ . (We factor out the singular term which is here $z − 3$ and expand the rest of the integrand in a Taylor series to get $z − 1$ coefficient.)

Since $sin ⁡ z$ goes as $z − z 3 6$ and $cos ⁡ z$ as $1 − z 2 2$ , $z cot ⁡ z$ behaves as $( 1 − z 2 2 ) ( 1 − z 2 6 )$ which can be written as $1 − z 2 6 − z 2 3 1 − z 2 6$ or $1 − z 2 3 ( 1 − z 2 6 )$ .
The residue of $cot ⁡ z z 2$ at $z = 0$ is therefore $− 1 3$ and we obtain the conclusion

$2 ∑ n > 0 ( n π ) − 2 − 3 = 0$

or

$∑ n > 0 n − 2 = π 2 6$

You can actually sum the first 128 (or 1024) terms of this sum on a spreadsheet and extrapolate by comparing the sum up to different powers of 2. If you extrapolate first forming $S 2 ( k ) = S ( 2 k ) − S ( 2 k − 1 )$ , then $S 3 ( k ) = 4 S 2 ( k ) − S 2 ( k − 1 ) 3$ then $S 4 ( k ) = 8 S 3 ( k ) − S 3 ( k − 1 ) 7$ , etc. You can get this answer to enormous accuracy numerically and verify this conclusion.