Complex Analysis

Question

Let P_N(x) be the polynomial given by

          P_N(x) = ∏^N_k=1 (x²+k²)

 

(a) For N≥1 evaluate  I_N = ∫^∞_-∞ (cos(πx))/(P_N(x))dx

 

(b) Show that I_N→0 as N→∞

 

 

Answer 1

Hi Lucy,

 

This is a residue calculus question, so you will want to find a complex-valued extension of the integrand or else a complex-valued integrand whose real part coincides with the integrand, and then integrate it over an appropriately chosen contour and do some estimates on the values of the integral on various parts of the contour and apply a limit process.

 

I can sketch the computation for the case N=1 and leave the general case to you.

 

I₁ = ∫_-∞^∞ cos(πx)/(x²+1) dx

 

will be replaced by 

 

I = ∫_C e^iπz/(z²+1) dz

 

where C is a semi-circular contour centered at the origin of radius R, whose diameter lies on the real axis, and which extends into the upper half plane (the upper half of a circle plus diameter).  

 

Note that the integrand, which I'll call f(z) from now on, has simple poles at z=±i, but only z=i lies within the contour.  Compute the residue at z=i:

 

Res(f,i) = lim_z→i (z-i)f(z) = lim_z→i e^iπz/(z+i) = e^-π/(2i).

 

Apply the Residue Theorem to find that

 

I = 2πi(e^-π/(2i)) = πe^-π.

 

Let Γ be the semi-circular part of the contour, and see that

 

I = ∫_-R^R f(x) dx + ∫_Γ f(z) dz

 

  = ∫_-R^R cos(πx)/(x²+1) dx + i ∫_-R^R sin(πx)/(x²+1) dx + ∫_Γ f(z) dz

 

  = ∫_-R^R cos(πx)/(x²+1) dx + ∫_Γ f(z) dz

 

  = πe^-π

 

where the observation that the integral involving sin(πx) is zero (odd function over symmetric interval) has been made.  

 

Now, take the limit as R→∞ and argue that the integral over Γ goes to zero (it's a standard estimate, might even be written out in your book), and you get finally that 

 

I₁ = πe^-π.

 

Of course, you will have to do this in the general case, which means computing the residues at all of the N poles enclosed by the contour, and observing that the resulting expression has a denominator that gets large as N tends to infinity (N! perhaps?).

 

Regards,

Hassan