Types of Real Integrals

Transcription

1 Math B: Complex Variables Types of Real Integrals p(x) I. Integrals of the form P.V. dx where p(x) and q(x) are polynomials and q(x) q(x) has no eros (for < x < ) and evaluate its integral along the fol- We will consider the complex function f() = p() q() lowing contour : which consists of the line L R ( < x < R) and of ĈR, the upper-half circle centered at the origin. Let R be large enough that all poles of f() in the upper-half plane (,..., n ) are located inside the closed contour. Then, using residue theory, f() d = πi k= Res f(). = k The value of the integral is found by splitting the integral over into two parts. In the limit as R, the integral over L R will give the desired real integral, and the integral over ĈR will go to ero. Following this plan, let α be the constant πi n k= Res = k f(), so the above equality becomes f() d + f() d = α. L R Then, parameteriing L R by = x ( < x < R), we have R f(x) dx + f() d = α.

2 Finally, we take the limit of both sides as R : f(x) dx + lim f() d = α. R All that remains is to show that CR b f() d. This is done by estimating the integral as follows: f() d ĈR max f() πr max f() C b R =R Since f() = p(), we estimate f() by using the triangle inequality to bound p(), but in q() order to bound, remember that we need q() larger than something! As an example, q() let p() = + and let q() = Using the triangle inequality, p() = + +, which equals R + on the circle = R. For the denominator, use the inequality , which equals R 6 3 on the circle = R (if R 6 > 3 ). This implies that on the circle. If our final bound for the integral is something that goes to R 6 3 ero as R goes to infinity that is, f() d F (R) as R then, the limit of the integral must be. (In the example above, F (R) = R +.) Finally, R 6 3 using this we have found the value of the real integral: f(x) dx = α. II. Integrals of the form P.V. f(x)sin(x) dx (or P.V. has no singularities on the real axis. f(x)cos(x) dx) where f(x) As before, we will integrate a complex function, but we won t simply replace x with. Instead, we consider the complex function g() = f() e i = f() cos() + if() sin(). Using the same contour (with R large enough that all poles of f() in the upper-half plane (,..., k ) are inside ), f()e i d = α where α = πi Res g(). = k Breaking up the integral over into two parts, we have R R f(x) cos(x) dx + i f(x) sin(x) dx + f()e i d = α. k=

3 Again, we only need to take the limit as R and show that CR b f()e i d. Then, matching up the real and imaginary parts of both sides, f(x) cos(x) dx = Re α and f(x) sin(x) dx = Im α. To show that the integral over the half-circle disappears in the limit, we often need Jordan s π inequality: e sin(θ) dθ < π (for R > ). If f is a polynomial with large enough power in R the denominator (so with enough decay at infinity), we can use the easier method of bounding the integral, as in II but if not, we require Jordan s inequality. For example, consider e i CR b d. Parameteriing the upper-half circle, we have + e i bcr + d = π (Re iθ )e ireiθ R e iθ + ireiθ dθ π R e sin θ dθ R πr R. In the last step, we used Jordan s inequality... without this extra decay from the integral, we can t prove that the right-hand side goes to ero! III. Integrals of the form π F(cos θ, sin θ) dθ, where F( + the unit circle., i ) has no poles on Notes: The integral may be taken over any interval of length π since the function is periodic. Also, you may see integrals where, for example, θ goes from to π by symmetry, these often equal one-half the value of the integral from to π! The integral in this case is really a parameteried version of a contour integral on the unit circle. Parameteriing the unit circle by (θ) = e iθ ( θ π or any other interval of length π!), we see that on the unit circle cos(θ) and sin(θ) can be written in terms of as Therefore, π cos(θ) = eiθ + e iθ = + F (cos θ, sin θ) dθ = F ( + C and, i sin(θ) = eiθ e iθ i ) d = πi k= =. i Res F ( +, ). = i k Make sure the residues in the sum are only for the poles inside the unit circle!! IV. Integrals that require different contours. Notes: These are integrals for which the complex function we want to integrate has a pole or a branch point somewhere on the real axis. In either cases, a contour that avoids going through the pole or the branch cut is needed! Examples:

4 sin x x dx = π This is similar to case II, except that the complex function g() = ei has a pole at =! Consider the following contour = L ( Ĉρ) L ĈR (where R > ρ > ): The function is analytic inside the contour, so for any R > ρ >, e i d =. Splitting the integral up, and parameteriing L by = r (ρ < r < R) and L by = r (ρ < r < R), R e ir ρ r ( dr) e i R bcρ d + e ir ρ r dr + e i d =. bcr e Using Jordan s inequality, we can show that lim i b R CR d =. Since = is a simple pole of g, we will use the Laurent series of g around to show that lim ρ bcρ e i The Laurent series of g() is b o + n= d = iπ Res = g(). e i bcρ d = b o bc ρ d + a n n d = b n= Cρ = iπ + a n n d. Cρ b n= a n n, where b o = Res g(). Then, = π iρe iθ ρe iθ dθ + a n n d Cρ b Simply show that the last sum tends to ero (notice that each of the integrals in the sum can be bounded by πρ n+!). Using these limits (notice that b o = in this case; also e ir e ir = i sin(r)), sin r i dr = iπ. r n=

5 dx = π x(x + ) Notice that x is simply the positive square root of a postive real number, but when we consider the complex function f() = specifiying the branch cut is necessary! ( + ) We shall use the branch < arg < π; therefore, f() is analytic away from the branch cut arg = and away from the pole at =. We will use the contour ɛ : We will take the limit ɛ before taking limits in ρ and R; therefore, it is fine to simplify by considering the contour as pictured on page 74 of your book where ɛ =. The interesting part is what happens on the two lines: L is parameteried by (x) = x+iɛ, and L is parameteried by (x) = x iɛ (with ρ ɛ x R ɛ of course, this will just become ρ x R when we take the limit ɛ.) L f() d = R ɛ ρ ɛ (x + iɛ) dx and (x + iɛ) + L f() d = R ɛ ρ ɛ (x iɛ) (x iɛ) + dx Make sure to use the branch cut when evaluating (x ± iɛ)! Since x + iɛ is in the first quadrant, but x iɛ is in the fourth, when we write these in polar coordinates using their arguments, we have Therefore, as ɛ, (x + iɛ) = x + ɛ e i arg(x+iɛ) x e as ɛ (x iɛ) = x + ɛ e i arg(x iɛ) x e πi as ɛ. f() d L R ρ R x (x + ) dx x e iπ f() d L ρ (x + ) dx. In the book, these are given as equalities since they assume the limit has already been taken that is, L is the line on the real axis, but with all limits being taken from above the line, and L is the line on the real axis with the limits taken from below. Finally, let ρ and R (and estimate to show that the integrals of f on both Ĉρ and ĈR go to ero!) ( ) dx = πi Res f() = πi ( ) = πi ( ) e i π = π x(x + ) =