Introduction to Residue Theorem for Definite Integrals
In this lesson, we explore how to use the residue theorem to evaluate definite integrals that involve sine and cosine functions. Integrals of the form (\int_0^{2\pi} F(\sin\theta,\cos\theta) d\theta) can be transformed into contour integrals over the unit circle in the complex plane. This process is closely related to Understanding the Residue Theorem in Complex Variables.
Using Complex Exponentials to Transform Integrals
- Recall that (z = e^{i\theta}) moves around the unit circle as (\theta) goes from 0 to (2\pi).
- Differentiate to find (d\theta = \frac{dz}{iz}).
- Use Euler's formula to rewrite sine and cosine:
- (\cos\theta = \frac{z + z^{-1}}{2})
- (\sin\theta = \frac{z - z^{-1}}{2i})
- Substitute these into the original integral to express it as a contour integral in terms of (z).
For foundational knowledge on these substitutions, see Mastering Trigonometric Identities, Equations, and the CAST Diagram.
Evaluating the Integral via the Residue Theorem
- The integral becomes a contour integral (\oint_C f(z) dz) over the unit circle.
- Identify singular points (poles) of (f(z)) inside the unit circle.
- Calculate the residue at each singularity:
- For simple poles, use (\text{Res}[f,z_0] = \lim_{z\to z_0} (z - z_0)f(z)).
- Sum the residues of all singularities inside the contour.
- Apply the residue theorem: (\oint_C f(z) dz = 2\pi i \times (\text{sum of residues})).
For techniques on residue calculation, refer to Mastering Residue Calculation Techniques in Complex Variables.
Example 1: Integral of (\frac{1}{5 + 4 \sin\theta}) from 0 to (2\pi)
- Transform to contour integral: [ \oint_C \frac{1}{5 + 2i (z - z^{-1})} \cdot \frac{dz}{iz} ]
- Find zeros of the denominator (poles):
- Use quadratic formula to find roots.
- Keep only the pole inside the unit circle.
- Compute residue at the pole inside the unit circle.
- Multiply residue sum by (2\pi i) to get the integral value: [ \int_0^{2\pi} \frac{d\theta}{5 + 4 \sin\theta} = \frac{2\pi}{3} ]
Example 2: Integral of (\sin^{2n} \theta) from 0 to (\pi)
- Use symmetry: (\int_0^{\pi} \sin^{2n} \theta , d\theta = \frac{1}{2} \int_0^{2\pi} \sin^{2n} \theta , d\theta).
- Express (\sin^{2n} \theta) in terms of (z) with complex exponentials.
- Use binomial theorem to expand ((z - z^{-1})^{2n}).
- Identify the constant term in expansion corresponding to the residue of the integrand.
- Apply residue theorem to compute the integral, resulting in: [ \int_0^{\pi} \sin^{2n} \theta , d\theta = \frac{(2n)!}{2^{2n}(n!)^2} \pi ]
Summary and Insights
- The residue theorem efficiently evaluates definite integrals involving sine and cosine by converting them to contour integrals.
- Using complex substitutions and Euler's formula simplifies the integration process, avoiding complicated real-variable techniques.
- Calculating residues at poles inside the unit circle provides a powerful shortcut for integral evaluation.
This approach not only solves otherwise difficult integrals but also deepens understanding of the relationship between complex analysis and real definite integrals. For a deeper theoretical foundation, consider reviewing Understanding Laurent Series and Residues in Complex Analysis and Understanding Cauchy’s Theorem and Complex Integrals Explained.
For further examples and to explore improper integrals, stay tuned for the next video.
welcome back everyone to another video on complex variables in this lesson I'm going to teach you how to use the
residue theorem to evaluate definite integrals in particular I'm going to look at definite integrals involving SS
and cosiness in the next video I'll move to improper integrals so let's begin in
this lecture I'm going to examine a particular class of integrals of this form where I'm integrating a function of
cosine Theta and sin Theta from 0 to 2 pi now because this integration involves the variable Theta varying from 0 to 2
pi you might immediately think of polar coordinates and circles when you look at an integral like this you might remember
from Polar coordinates that Theta is used to represent the angle relative to the positive xais and you might also
remember that Theta varies from 0 to 2 pi so in varying Theta from 0 to 2 pi for a constant radius we're essentially
going around a circle in fact you could say that this integral essentially involves integrating the function
capital F along a unit circle with Theta representing the angle in radians from the positive X AIS which varies from 0
to 2 pi we can actually take advantage of the similarity and make a change of variables we can use the polar
representation of complex numbers to let our complex number Z equal the exponential of I * Theta then we can
make the Rel relevant substitutions from Theta to Z we can replace the differential of theta by the
differential of Z by taking the derivative of equation one with respect to Theta in that case the dzd Theta is
just I * the exponential of I Theta which can be Rewritten as I * Z just by plugging in the Z from equation 1 so
just switch around the differentials and we'll find that D Theta is just dz/ I Y Z now all that's left is to switch the
cosine Theta and sin Theta to functions of Z and that's actually pretty easy we're going to use Oiler formula to
expand equation one in terms of ss and cosiness we can also write another equation involving 1 / Z which would
just be the exponential of I * Theta expanding this out using Oilers formula will yield cine Theta minus I sin Theta
I'll just call this equation 3 now let's add equations 2 and 3 and get an expression for cosine Theta in terms of
Z we can also subtract 2 and three to get an expression for sin Theta in terms of
Z so if we plug all these Expressions into our integral up here then we'll end up with the Contour integral over C of
the function capital F now all in terms of Z time DZ over i z note that the C here the curve you're integrating over
is the unit circle centered at the origin now once you convert your integral in terms of theta into an
integral in terms of Z you can then use the residue theorem to evaluate the integral so let's do an example
involving finding a definite integral that contains s and cosine in this example we want to integrate 1 5 + 4 sin
Theta from 0 to 2 pi this is where you'll see why the residue theorem is so useful because if I were to ask you to
integrate this using the regular techniques you learned in calculus 2 then you wouldn't be able to do anything
here regular substitution won't work partial fractions won't work and integration by parts won't work you'll
have to use a technique called The Wire St substitution which isn't even covered in most calculus 2 courses it's a rather
exotic technique but using the residue theorem however does allow you to solve this integral in a manner that you might
argue is even more convenient than the y or Strat substitution so let's get started notice that this integral
follows the form up here so we can immediately move to substituting Z in place of theta and converting this into
a contour integral over the unit circle if we do that then our integral becomes the Contour integral over the unit
circle C of 1 over 5 + 4 * Z - Z inverse / 2 I which is just sin Theta as we showed earlier time DZ i z which is just
the differential D Theta reexpressed in terms of DZ as we showed earlier as well we can then simplify this integral and
get the following expression what we want to do now is use the residue theorem to evaluate this
integral but before we use the residue theorem let's take the time to recall the exact statement the exact statement
here is that if F of Z is some complex function with a bunch of singular points or poles then the Contour integral of f
of Z over a simple closed curve C which encloses all these poles is just 2 pi I times the sum of the residues of f of Z
at each of the enclosed singular points now how is the residue theorem going to help us for this example well
here we're integrating this function that I'll conveniently label F of Z over a unit circle which is our simple closed
Contour C so if I need to evaluate this integral I can find the singular points of f of Z find their corresponding
residu add them together and use the result to determine the value of the integral via the residue
theorem so let's start applying the residue theorem by first finding the singular points of our function the
singular points here would be the values of Z at which the function is undefined which caus the denominator to be zero to
find those singular points I can use the quadratic formula to find the zeros of the
denominator if I do that then I'll get two distinct values of Z which make the denominator zero the first is going to
be 0.5 I if I use the plus sign in the quadratic formula and the second is - 2i if I use the minus sign in the quadratic
formula now this first root clearly has a magnitude less than one so it's going to be inside our unit circle however the
second root has a magnitude greater than one so it's going to be outside our unit circle we're only concerned about the
residues for the poles inside the Contour of integration so we only need to find the residue at
Z1 now let's rewrite this integral in factored form since we've already found the zeros of the
denominator and let's now find the residue at Z1 since Z minus Z1 or Z + 0.5i only appears in the denominator
once it clearly corresponds to a simple Pole if you remember the previous lecture
then we can find the residues at simple poles by multiplying F of Z by Z minus Z1 and taking the limit as Z approaches
Z1 and if you compute the limit you'll find that the residue is just 1 over 3 I and this is the only residue we're
concerned with if we now apply the residue theorem we'll find that the integral over C of dz/ 2 * Z + 2 I * Z +
0.5 I = 2i I * 1 3 I which = 2 piun 3 so therefore we can conclude that the integral from 0 to 2 piun of 1/ 5 + 4
sin Theta is just 2 piun / 3 and that does it for this example pretty simple stuff no ridiculous wire stess
substitution necessary let's now do another example here we're going to find the integral
from 0 to Pi of sin Theta raised to^ 2 N where n is some posi integer the issue here is that this
integral is from 0 to Pi but from what we've learned so far we want our integrals to be between 0 and 2 pi but
going around this issue isn't too difficult once you recognize that sin Theta raised to the power 2 N is the
same from PI to 2 pi as it is from 0 to Pi go ahead plot the graph for a bunch of NS and you'll see that this is
correct this fact is actually pretty convenient because we can exploit the symmetricity to say that the integral
from 0 to Pi is just half the integral from 0 to 2 pi since the function being integrated repeats itself once we reach
Theta equal Pi so now that the integral is in the proper form meaning from 0 to 2 pi we
can use the substitutions to convert Theta to Z and express our integral purely in terms of Z when we convert
everything to Z this is what we'll end up with note that c is again just the units Circle centered at the
origin this looks like a pretty complicated expression but we can simplify it take out the 2 I so you end
up with 2 i^ 2 N now the 2 ^ 2 N can just be written as 4 the N while the I to the^ 2 N can be written as -1 to the^
n let's take out all the constants from the integral and leave only the terms involving Z Now the Z minus Z inverse to
the 2 N term seems pretty difficult to manage but we can deal with it using the binomial theorem the reason we want to
use the binomial theorem is that using the binomial theorem will ultimately give us an expansion of this expression
which we can use to find the residue of this integrand which we'll use to apply the residue theorem so let's recall the
binomial theorem which states that a + b^ m is the sum from Ral 0 to M of a the M - R * B to the r * m choose r
so if we apply the binomial theorem to the numerator of our integrant to the numerator of the function we're
integrating we'll find that the k + 1 term of this expansion is Z 2 nus K * -1 K * Z
inverse to the^ K * 2 N chose K let's simplify this by combining the terms involving Z and we'll get the
following and now let's substitute this into the integral but now will have a summation out front because a bunch of
these terms have to be added together to give the binomial expansion of Z minus Z inverse to the^ n now what we want here
is the term corresponding to the residue term in the luron expansion of the integral since we're already dividing by
Z in the expression for the integral all we have to do here is find the term in the binomial expansion for the numerator
which is a constant because the constant divided Z will just be the residue term we can see that the constant term in
this binomial expansion will be the term where k equals n so the coefficient of the term with k equal n is then the
residue of the integrant around the pole at zal 0 which happens to be the only poll we're dealing with here that means
the residue of this integrand is just -1 n * 2N factorial Over N factorial s and this also happens to be the only residue
since there's only one pole inside the curve C and that's zal 0 so now if we apply the residue theorem to compute
this whole integral this is what we'll get and if we simplify this expression we'll finally end up with 2 N factorial
over 2^ 2 n * n factorial 2 * PK so we found the integral of s of theta to the^ 2N using the residue theorem and that
should do it for this video If you enjoyed the lesson feel free to like And subscribe in the next complex variables
video I'm going to be using the residue theorem to compute improper integrals thank you for watching and this is the
faculty of Khan signing out
The residue theorem can be applied by converting a definite integral involving sine and cosine, such as (\int_0^{2\pi} F(\sin\theta, \cos\theta) d\theta), into a contour integral around the unit circle in the complex plane. Using the substitution (z = e^{i\theta}), sine and cosine are rewritten in terms of (z) via Euler's formula, enabling the integral to be expressed as (\oint_C f(z) dz). Then, by identifying and calculating the residues of poles inside the unit circle, the integral value is found using the residue theorem formula (2\pi i \times) sum of residues.
Start by substituting (z = e^{i\theta}) which traces the unit circle as (\theta) goes from 0 to (2\pi). Then express (\sin\theta) and (\cos\theta) using their complex exponential forms: (\cos\theta = \frac{z + z^{-1}}{2}) and (\sin\theta = \frac{z - z^{-1}}{2i}). Replace (d\theta) with (\frac{dz}{iz}). This transforms the original integral into a contour integral over (z) along the unit circle, making it suitable for complex analysis methods like the residue theorem.
Identify singularities (poles) of the transformed integrand (f(z)) by solving where the denominator equals zero. Then determine which poles lie inside the unit circle (i.e., have magnitude less than 1). For simple poles, compute the residue using (\text{Res}[f,z_0] = \lim_{z \to z_0} (z - z_0)f(z)). Summing these residues allows application of the residue theorem to evaluate the contour integral, which corresponds to the original definite integral.
Yes. Consider (\int_0^{2\pi} \frac{d\theta}{5 + 4 \sin\theta}). By substituting (z = e^{i\theta}) and expressing (\sin\theta) in terms of (z), the integral becomes (\oint_C \frac{1}{5 + 2i (z - z^{-1})} \cdot \frac{dz}{iz}). By finding poles via the quadratic formula and selecting the one inside the unit circle, calculate its residue and multiply by (2\pi i) to get (\frac{2\pi}{3}), the value of the original integral.
Symmetry allows us to relate the integral over ([0, \pi]) to half the integral over ([0, 2\pi]): (\int_0^{\pi} \sin^{2n} \theta , d\theta = \frac{1}{2} \int_0^{2\pi} \sin^{2n} \theta , d\theta). Expressing (\sin^{2n} \theta) in terms of (z) using complex exponentials and then expanding with the binomial theorem helps identify the term that corresponds to the residue. Applying the residue theorem yields the exact value (\frac{(2n)!}{2^{2n}(n!)^2} \pi).
Complex analysis, through the residue theorem and contour integrals, transforms challenging real-variable integrals into more manageable contour integrals in the complex plane. Using complex exponentials simplifies trigonometric expressions, avoids cumbersome real calculus techniques, and leverages the powerful residue theorem to evaluate integrals efficiently and elegantly.
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