Introduction to Improper Integrals
Improper integrals extend the concept of definite integrals to infinite or unbounded intervals.
- Typically evaluated by replacing infinite limits with parameters and taking limits as those parameters approach infinity.
- Convergence depends on these limits existing; otherwise, the integral diverges.
Cauchy Principal Value
When an improper integral diverges, the Cauchy principal value (CPV) can sometimes assign a finite value:
- Defined by symmetrically extending integration limits to ±R and taking the limit as R approaches infinity.
- CPV may exist even if the improper integral itself diverges.
- For even functions, if CPV exists, the improper integral converges and equals the CPV.
Theorem for Even Functions
If f(x) is even (f(x) = f(−x)) and the CPV exists, then:
- The improper integral from −∞ to ∞ converges.
- The integral equals the CPV.
- Over semi-infinite intervals, the integral equals half the CPV.
Using the Residue Theorem to Evaluate Improper Integrals
The residue theorem from complex analysis provides a powerful method to compute these integrals under certain assumptions; see Understanding the Residue Theorem in Complex Variables for foundational concepts.
Assumptions
- f(x) is a rational function P(x)/Q(x) with real coefficients.
- P and Q have no common factors.
- The degree of Q is at least two greater than the degree of P.
- Q(z) has no real zeros but has zeros in the upper half of the complex plane.
Procedure
- Identify Poles: Find all zeros of Q(z) with positive imaginary parts.
- Contour Setup: Select a semicircular contour in the upper half-plane, combining the real axis segment (−R to R) and a semicircular arc of radius R.
- Apply Residue Theorem: The integral over the closed contour equals 2πi times the sum of residues at poles inside the contour.
- Evaluate Semicircular Integral: Show this integral tends to zero as R → ∞ using polar representation (z = Re^{iθ}).
- Conclusion: The principal value of the integral over the real axis equals 2πi times the sum of residues.
If f(x) is even, then this principal value equals the improper integral.
For mastery in calculating residues, refer to Mastering Residue Calculation Techniques in Complex Variables.
Example: ∫_{−∞}^{∞} 1/(x2 + 1) dx
- f(x) = 1/(x2 + 1) meets all assumptions:
- Rational function with real coefficients.
- Degree of denominator is 2 greater than numerator.
- Denominator has no real zeros.
- Poles: z = i (upper half-plane) and z = −i (lower half-plane).
- Contour integral setup and residue calculation at z = i:
- Residue found using limit formula: Res = 1/(2i).
- Semicircular integral → 0 as R → ∞.
- Integral evaluates to 2πi * (1/2i) = π.
- f(x) is even, so improper integral converges and equals π.
This matches the answer obtained via trigonometric substitution, validating the method.
For deeper understanding, see Using the Residue Theorem to Evaluate Definite Integrals Involving Sine and Cosine which covers related integral evaluations.
Summary
- The residue theorem provides an elegant technique for evaluating certain improper integrals.
- Key conditions on the function and contour ensure the contributions from arcs vanish.
- The Cauchy principal value plays a critical role, especially for even functions.
- Practical residue computation yields explicit integral values.
Additional Resources
- How to find residues: [link provided in original video]
This method bridges calculus and complex analysis, offering a versatile approach for advanced integral evaluation.
For background on series expansions and residues, see Understanding Laurent Series and Residues in Complex Analysis and to connect foundational theory, refer to Understanding Cauchy’s Theorem and Complex Integrals Explained.
greeting students and welcome back to another lecture on complex variables in this video I'm going to show you how to
use the residue theorem to evaluate improper integrals so let's start by reviewing basic calculus and let's look
at how improper integrals are usually evaluated using calculus techniques recall that if I had an improper
integral of f of X over a semi-infinite intervals so from 0 to infinity I could evaluate it by replacing the
infinite upper bound by some parameter R and then taking the limit as R approaches infinity when this limit
exists we can say that the improper integral converges and if this limit doesn't exist and the improper integral
diverges similarly if I had an improper integral over an infinite interval I could evaluate that integral by
replacing the infinite bounds by two parameters R 1 and R 2 then splitting up the integral and then taking the limits
of each of those two split up integrals as R 1 and R 2 approached infinity in this scenario both limits have to exist
for the improper integral to converge by the way I'm gonna call this equation 1 but what if by improper integral
diverged does that mean I should start crying drop out of school and ask my student loans to be forgiven of course
not because student loans can't be forgiven but staying on topic even if the improper integral diverges we can
still assign a value to it in some cases using something called the couchy principal value the couchy principal
value of an improper integral from negative infinity to infinity of f of X DX is found by replacing the infinite
upper and lower bounds by some parameter R and then taking the limit of the entire integral as capital R approaches
infinity I'm gonna call this equation 2 now in some cases the couchy principal value of an integral exists even when
the integral diverges take for example the integral of x over an infinite interval if we find the improper
integral using equation 1 this first term will approach negative infinity and the second term will approach positive
infinity since the sum of negative infinity and positive infinity is undefined this
integral would diverge however if we find the couchy principal value of the integral using equation two the answer
comes out to zero so the couchy principal value exists even though the improper integral itself diverges and so
this is how the couchy principal value can be used to assign numbers to integrals which would otherwise diverge
now an interesting fact about the couchy principal value is that when the improper integral does exist the couchy
principal value equals that improper integral we can prove this very easily starting with equation two if we split
up this integral along the right into two parts and then if we distribute the limit over the integrals we end up with
an expression that equals the improper integral in equation one as long as these two limits exist of course
now there's another theorem that we're going to prove here before I show you how to use the residue theorem to
evaluate improper integrals this theorem concerns even functions and it says that if f of X is an even function and if the
couchy principal value of the integral of f of X from negative infinity to infinity exists then the improper
integral itself converges and equals the couchy principal value so for even functions the couchy principal value is
equal to the improper integral as long as the couchy principal value exists so let's prove this theorem because f of X
is even it's integral from negative R 1 to 0 equals 1/2 the integral from negative R 1 to R 1 just because of the
symmetry inherent in an even function an even function is where f of X equals F of negative x in other words that
function reflects evenly about the y axis similarly the integral from 0 to R 2 equals 1/2 the integral from negative
R 2 to R 2 because f of X is even now when we add these two equations here's what we'll end up with if we take the
limit of both sides as R 1 and R 2 approach infinity will find that the left-hand side matches the definition of
the improper integral in an equation 1 while the right hand side just equals two halves of the principal value added
together and since we know that the principal value exists from the assumption of the proof the left-hand
side or the improper integral must also exist because the left and the right are equal to each other and as a result we
can conclude that the improper integral indeed equals the principal value when f of X is even also if we just take the
second equation only in R 2 and let R to approach infinity then we'll find that the improper integral over a semi
infinite interval of an even function equals 1/2 the principal value if the principal value exists I'm going to call
this guy equation 3 and the sky' equation for any way that does it for our preamble
it's time to get to the fun part how do we compute improper integrals the other residue theorem well we'll start with
four assumptions that we need to make before we can actually begin the procedure
suppose we had a function f of X that was a rational function so the ratio of two polynomials P and Q suppose also
that P and Q have real coefficients and no common factors then suppose that the degree of Q is at least two greater than
the degree of P and finally suppose that Q of Z has no real zeros but at least one zero about the real axis if all four
of these conditions hold then we can use a special technique to evaluate the improper integral of f of X over an
infinite interval the first step of this technique is to evaluate all the zeroes of Q of Z that are above the real axis
in other words we first need to find all the zeros that have a positive imaginary part the second step of this technique
is to set up a closed semicircular contour which will be integrating f of Z over I'm gonna call that closed curve C
now this semicircular region will have a radius of R and the curve C enclosing the semicircular region can be divided
into two parts the first part which I'll draw in white will be a straight line from negative capital R to capital R on
the real axis the second part which I'll call C sub R will be a semicircular arc on the circle which I'll draw over in
blue now if we apply the residue theorem then the integral of f of Z over the entire contour C is just 2 pi I times
the sum of the residues of f of Z at the poles which are inside the contour C in addition the integral of f of Z over the
entire contour equals the integral over the real axis plus the integral over the semicircular arc so if we substitute
that in here's what we'll get now if we replace Z in this first integral by the real number X which doesn't really
change anything since the first integral is over the real axis anyway and if we then take the limit of this entire
equation as capital R approaches infinity will find that this first integral becomes the principal value of
the integral of f of X over an infinite interval of course this is according to the definition of the principal value
and this is equal to the limit of two pi I times to sum up the residues of f of Z minus the limit of the integral of f of
Z over the semicircular contour the fourth step of this technique is to evaluate the semicircular contour
integral now this step is an as difficult as it might seem because along a semicircle if radius R the magnitude
of the complex number Z or the distance of Z from the origin is a constant equal to capital R so what we can do is use
the polar representation of complex numbers to write Z as R times the exponential of I theta this will allow
us to write DZ in terms of D theta and convert our semicircular integral over Z into an integral over the angle theta
and now since things are in terms of capital R more explicitly we can easily take the limit of the Steeda integral as
capital R approaches infinity usually this will turn out to be 0 because of the conditions we stipulated earlier on
but it's still a good idea to make sure and it's also good if your f of Z doesn't entirely obey the conditions
that we stipulated earlier on specifically the condition where the degree of Q has to be 2 greater than the
degree of P once we're done with step four and once we've found all the residues we can then find the principal
value of the improper integral using this equation that we wrote down I've gotten rid of the limit on the residues
by the way because the residues don't depend on capital R now if f of X is an even function then we can replace the
principal value by the improper integral itself according to the second theorem that we proved above and this will allow
us to find the improper integral of f of X over an infinite interval by the way if we wanted to find the improper
integral over a semi infinite interval we would just have this whole equation so in the end what we're getting from
our four conditions is just the principle value but if f of X is an even function then we can find the improper
integral directly for a function that isn't even we would need to make sure that its improper integral converges
over our infinite or semi infinite interval anyway that should cover the technique for computing improper
integrals by the residue theorem it looks like a lot of work is involved but let's put everything in the context by
doing a quick example on the side first in this example we want to find the improper integral of 1 over x squared
plus 1 from the four conditions we wrote earlier on we can see that f of X is a rational function the coefficients are
all real and there's no common factor we can also see that the degree of the denominator is two greater than the
degree of the numerator x squared compared to X to the power 0 and finally we can see that the denominator has no
real zero so we can go ahead with our technique the first step is to find the zeros of Q of Z which is pretty easy
since we can factor this into Z plus I and Z minus I again I squared is negative 1 I is the imaginary number the
0 above the real axis here is obviously just Z equals I in the second step we set up our contour integral which
consists of an integral over the real axis plus an integral over a semicircular arc C sub R and in the
third step we apply the residue theorem and take the limit as capital R approaches infinity to get the principal
value of the improper integral now that we're done the third step we can move on to the fourth step in which we evaluate
the integral over the semicircular arc as R approaches infinity since the magnitude of Z is R over the
semicircular arc we can use the polar representation of complex numbers to write this integral in terms of theta
now if we take the limit as capital R approaches infinity this integrand approaches zero because there's an R
squared in the denominator and only one R in the numerator so this entire integral over the semicircular arc is 0
because the definite integral of 0 is just 0 so now we're left with step 5 there's
only one poll we're concerned with and that's at Z equals I so if we want to find the residue of 1 over 1 plus Z
squared at Z equals I which happens to be a simple pole here because Z minus I only appears once in the denominator
what we can do is multiply by Z minus I and take the limit as Z approaches I we already covered this in my how to find
residues video by the way I put the link in the description anyway after doing a bunch of computation will find that the
residue of 1 over Z squared plus 1 at Z equals I is just 1 over 2i and since this is the only residue we're concerned
with we can plug it back into the residue theorem equation and conclude that the principal value of our improper
integral is 2 pi I times 1 over 2i which is just PI and finally because 1 over x squared plus 1 is an even function
because of the x squared because it's an even function we can conclude that the improper integral converges and it's
equal to the principal value which means that the improper integral of 1 over x squared plus 1 over an infinite interval
is PI so we've successfully evaluated an improper integral using the residue theorem just one thing to note before I
wrap up you could evaluate this integral using another method of trig substitution the antiderivative of the
integrand here is just tan inverse so if you apply the limits you'll find pi as your answer which is consistent with
what we found using the residue theorem so it's a nice verification anyway that should do it for my lecture I just like
to finish off by thanking my patrons Jacob Soares and Jennifer Hoffman for donating at the five-dollar level or
higher to my patreon if you would like to become a patron I put a link to my patreon account in the description and
you can support me there if you want so that's it if you enjoyed the video feel free to like and subscribe this is the
Faculty of Khan signing out
An improper integral extends the concept of definite integrals to cases where the integration interval is infinite or the integrand has unbounded behavior. Unlike regular definite integrals with finite limits and bounded functions, improper integrals are evaluated by introducing limits that approach infinity or the points of discontinuity and checking if these limits exist to determine convergence.
The Cauchy principal value (CPV) assigns a finite value to some improper integrals that diverge by symmetrically extending the limits from -R to R and taking the limit as R approaches infinity. This method can provide meaningful values even when the integral itself does not converge traditionally, particularly for even functions where CPV existence implies the improper integral converges to the same value.
To apply the residue theorem effectively, the function should be a rational function P(x)/Q(x) with real coefficients where P and Q share no common factors. Additionally, the degree of Q must be at least two greater than P, Q(z) should have no real zeros, and all zeros of Q(z) relevant for evaluation lie in the upper half of the complex plane. These conditions ensure the contour method and limit processes yield correct integral values.
Yes. First, identify all poles of the integrand in the upper half-plane. Next, construct a semicircular contour comprising the real axis segment from -R to R and the semicircle of radius R in the upper half-plane. Apply the residue theorem stating the contour integral equals 2πi times the sum of residues at enclosed poles. Then, show the integral over the semicircular arc tends to zero as R approaches infinity. Finally, conclude that the improper integral's principal value equals 2πi times the sum of residues.
Having the denominator’s degree at least two greater than that of the numerator ensures that the integrand decays sufficiently fast at infinity. This decay makes the integral over the semicircular arc of the contour approach zero as the radius goes to infinity, a crucial step in validating that the contour integral equals the improper integral over the real axis.
For f(x) = 1/(x² + 1), the poles are at z = i and z = -i, with only z = i in the upper half-plane. Calculating the residue at z = i gives 1/(2i). The contribution from the semicircular arc tends to zero, so the integral equals 2πi times the residue (2πi * 1/(2i)) = π. Since f(x) is even, the improper integral converges and equals π, matching classical integral results.
For even functions where f(x) = f(−x), if the Cauchy principal value exists, the improper integral from -∞ to ∞ converges and equals this principal value. This relationship leverages the symmetry of the function and the definition of CPV to handle integrals that might otherwise be difficult to evaluate through direct limits.
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