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UNIVERSITY OF BRISTOL
Department of Physics Level 3 Mathematical Methods 33010 (2009/2010)
Definite Integrals by Contour Integration
Many kinds of (real) definite integrals can be found using the results we have found for contour integrals in the complex plane. This is because the values of contour integrals can usually be written down with very little difficulty. We simply have to locate the poles inside the contour, find the residues at these poles, and then apply the residue theorem. The more subtle part of the job is to choose a suitable contour integral i.e. one whose evaluation involves the definite integral required. We illustrate these steps for a set of five types of definite integral.
Type 1 Integrals
Integrals of trigonometric functions from 0 to 2π:
I =
∫ (^2) π
0
(trig function)dθ
By “trig function” we mean a function of cos θ and sin θ. The obvious way to turn this into a contour integral is to choose the unit circle as the contour, in other words to write z = exp iθ, and integrate with respect to θ. On the unit circle , both cos θ and sin θ can be written as simple algebraic functions of z:
cos θ =
(z + 1/z) sin θ =
2 i
(z − 1 /z)
and making this replacement turns the trigonometric function into an algebraic function of z whose poles can be easily found. Example:
I =
∫ (^2) π
0
dθ 1 + a cos θ
where − 1 < a < +
I =
1 + a 2 (z + (^1) z )
dz iz
i
∮ (^) dz
2 z + az^2 + a
The poles of the function being integrated lie at the roots of the equation az^2 + 2z + a = 0 i.e. at the points
z± =
a
( − 1 ±
1 − a^2
)
C
Z- Z+
Of the poles, only z+ lies inside the unit circle, so I = 2πiR+ where R+ is the residue at z+ To find the residue we note that this is a simple pole and if we write the integrand as f (z) = g(z)/h(z) the residue at z+ is:
g(z+) h′(z+)
2 i(az+ + 1)
i
1 − a^2
Hence the integral required is 2π/
1 − a^2
Type 2 Integrals
Integrals such as I =
∫ (^) +∞
−∞
f (x)dx
or, equivalently, in the case where f (x) is an even function of x
I =
∫ (^) +∞
0
f (x)dx
can be found quite easily, by inventing a closed contour in the complex plane which includes the required integral. The simplest choice is to close the contour by a very large semi-circle in the upper half-plane. Suppose we use the symbol “R” for the radius. The entire contour integral comprises the integral along the real axis from −R to +R together with the integral along the semi-circular arc. In the limit as R → ∞ the contribution from the straight line part approaches the required integral I, while the curved section may in some cases vanish in the limit. Note: when we
C 3 C 1
C 2
-R R
choose to split up the complete contour into sections ( C 1 , C 2 , etc.) we shall use the notation (J 1 , J 2 , etc.) for the corresponding contour integrals. i.e.
J 1 =
∫
C 1
f (z)dz etc.
and since in polar coordinates eiz^ = eir^ cos^ θe−r^ sin^ θ^ the numerator tends to zero as r becomes large everywhere in the upper half-plane where sin θ is positive. Using the same D-shaped contour as before, the semi-circular arc contributes
∫
arc
eiz^ dz z^2 + a^2
= lim R→∞
∫ (^) +π
0
eiz^ iReiθdθ R^2 e^2 iθ^ + a^2
So I = 2πiΣR where the sum is of the residues in the upper half-plane. The function has simple poles at z = ±ia of which z = +ia is in the upper half-plane with residue R = (z − ia)f (z)|z=ia = e
iz z+ia |z=ia^ =^ −^
i 2 a e
−a (^) So finally, taking real parts, I = π a e
−a.
This argument will clearly work whenever the integral around the semi-circle van- ishes. The previous discussion shows this is true for functions of the form
f (z) =
P (z) Q(z)
eiz
where P, Q are polynomials in z and the order of Q is at least 2 greater than that of P , because in the limit of large R, the contribution from the arc will fall as 1/R ( or a higher inverse power ). If, on the other hand, the order of Q is just one higher than that of P , it is not immediately clear what will happen. However, in such cases the exponential decrease of e−R^ sin^ θ^ in the upper half-plane overwhelms everything else and the arc integral still vanishes. This fact, which we do not have time to prove formally, is known as “Jordan’s Lemma” and it makes contour integration a useful method for a large class of integrals, and you should know it and be ready to use it in appropriate cases.
Type 4 Integrals
A type of integral which brings in some new ideas is similar to Type 2 but with a pole of the integrand actually on the contour of integration. As an example of a situation where this arises, consider the real integral
I =
∫ (^) +∞
−∞
cos x x
dx
The approach previously discussed would involve replacing cos x/x by eiz^ /z , in which case the semi-circular arc would vanish by Jordan’s lemma. However there is a problem, because cos x/x has a pole at the origin, and the integrand diverges as we approach the origin along the real axis either from positive or negative values of z. To deal with this we introduce the concept of the “Principal Value” of a definite integral. In this we imagine that we exclude a small symmetrical range of real z values around the pole, and take the limit as this excluded region shrinks to zero width. We can find this by a suitable contour integral. To do this in our example we find the contour integral of eiz^ /z around a contour similar to that used above, but also involving a small semi-circular detour around the pole at the origin: There are no poles inside this contour so the total contour integral vanishes ( J 1 + J 2 + J 3 + J 4 = 0 ). The integral around the big semi-circle (J 2 ) also vanishes in the limit of large radius, and the integral along the real axis
C 3 C 1
C 2
C 4
(J 3 + J 1 ) is what we have just defined as a Principal Value in the limit as r → 0 and R → ∞. To find the contribution from the small semi-circle (J 4 ) we evaluate
J 4 = lim r→ 0
∫ (^0)
−π
eire iθ ireiθ^ dθ reiθ^
= i
∫ (^0)
−π
1 dθ = −πi
The vanishing of the entire contour integral yields
0 = J 1 + J 2 + J 3 + J 4 = P V
∫ (^) +∞
−∞
eixdx x
So we can separate real and imaginary parts to obtain
P V
∫ (^) +∞
−∞
cos x x
dx = 0
∫ (^) +∞
−∞
sin x x
dx = π
(We have omitted the “PV” in the final integral because sin x^ x is actually finite at x = 0.)
Type 5 Integrals
Our last type of integral will be those involving branch cuts. Far from being a problem, these can actually make some kinds of definite integral possible because we can make use of the discontinuity across the cut to construct the required integral. This is best shown by an example:
Example I =
∫ (^) +∞
0
dx x^3 + 1
resembles Type 2, but because the integrand is not even we cannot extend the integration to the whole real axis and then halve the result. However, suppose we look at the contour integral
J =
∫
C
ln zdz z^3 + 1
around the contour shown. Note that this contour does not pass through the cut onto another branch of the function. Remember that ln z = ln r + iθ + 2πin where n is an integer distinguishing the branches of the function. On our contour we have points just above the cut in the section C 1 and points at the same x values but just below the cut in C 3. Because we stay on the same sheet (say n = 0) throughout our contour, these values differ by 2πi. The sections C 1 and C 3 are described in opposite
