an Advanced Mathematics question?

Question

I can not prove the followng:

sign of integration(-infinite,infinite)
{[cos(x)/(x-w)]dx};
w----a constant;

the answer is
pi*i*e^(i*w) when Im*w>0
-pi*i*e^(-i*w) when Im*w<0

can anybody help,show me the detailed all the proof steps?
my email is:[e-mail]

Answer 1

[-∞, ∞]∫cos x/(x-w) dx

First, we will rewrite this improper integral formally as a limit:

[a→∞]lim [-a, a]∫cos x/(x-w) dx

Next, we will rewrite cos x as (e^(ix) + e^(-ix))/2. We will also change the variable of integration to z to remind ourselves that we are working with complex numbers:

1/2 [a→∞]lim [-a, a]∫(e^(iz) + e^(-iz))/(z-w) dz

Rewriting this as the sum of two integrals:

1/2 [a→∞]lim [-a, a]∫e^(iz)/(z-w) dz + [-a, a]∫e^(-iz)/(z-w) dz

Now, let us consider the integrals one at a time. First let us find:

[a→∞]lim [-a, a]∫e^(iz)/(z-w) dz

We will compute this by computing the value of a related contour integral. Let γ₁ be the straight line from -a to a, γ₂ be the straight line from a to a + ai, γ₃ be the straight line from a+ai to -a + ai, and γ₄ be the straight line from -a + ai to -a. So the path γ₁ + γ₂ + γ₃ + γ₄ traces out a large rectangle in the upper half-plane. If the point w does not appear inside this rectangle, then e^(iz)/(z-w) is analytic on the inside of the rectangle, and [γ₁ + γ₂ + γ₃ + γ₄]∫e^(iz)/(z-w) dz = 0. If w IS on the inside of the rectangle, then e^(iz)/(z-w) has a simple pole at w, with a residue of e^(iw), so by the residue theorem [γ₁ + γ₂ + γ₃ + γ₄]∫e^(iz)/(z-w) dz = 2πi*Res(e^(iz)/(z-w)) = 2πie^(iw). Since, if Im(w) is positive, we will have w in the rectangle for all sufficiently large values of a, and if Im (w) is negative, we never will, we have that:

[a→∞]lim [γ₁ + γ₂ + γ₃ + γ₄]∫e^(iz)/(z-w) dz = {0 if Im(w) < 0, 2πie^(iw) if Im (w) > 0}

Now we will break the integral into its four components and show that three of them go to zero. First, parameterize γ₂ by letting γ₂(x) = a+ix for 0≤x≤a. Then we have:

[a→∞]lim |[γ₂]∫e^(iz)/(z-w) dz|
= [a→∞]lim |[0, a]∫e^(i(a+ix))/(a+ix-w) i dx|
≤ [a→∞]lim [0, a]∫|e^(ia-x)/(a+ix-w) i| dx
= [a→∞]lim [0, a]∫|e^(ia-x)|/|a+ix-w| dx
= [a→∞]lim [0, a]∫e^(-x)/|a+ix-w| dx
≤ [a→∞]lim [0, a]∫e^(-x)/||a+ix| - |w|| dx
≤ [a→∞]lim [0, a]∫e^(-x)/(a - |w|) dx (since for sufficiently large values of a, a>|w|, so 0 < a-|w| ≤ |a+ix| - |w| = ||a+ix| - |w||)
= [a→∞]lim 1/(a - |w|) [0, a]∫e^(-x) dx
= [a→∞]lim 1/(a - |w|) (-e^(-a) + 1)
= 0

So since [a→∞]lim |[γ₂]∫e^(iz)/(z-w) dz| = 0, it follows that [γ₂]∫e^(iz)/(z-w) dz = 0 as well. Similarly, parameterizing γ₄ with γ₄(x) = -a+i(a-x), 0≤x≤a, we have:

[a→∞]lim |[γ₄]∫e^(iz)/(z-w) dz|
= [a→∞]lim |[0, a]∫e^(i(-a+ia-ix))/((-a+ia-ix)-w) (-i) dx|
≤ [a→∞]lim [0, a]∫|e^(i(-a+ia-ix))/((-a+ia-ix)-w) (-i)| dx
= [a→∞]lim [0, a]∫e^(-a+x)/|-a+i(a-x) - w| dx
≤ [a→∞]lim [0, a]∫e^(-a+x)/||-a+i(a-x)| - |w|| dx
≤ [a→∞]lim [0, a]∫e^(-a+x)/(a - |w|) dx
= [a→∞]lim 1/(a-|w|) [0, a]∫e^(-a+x) dx
= [a→∞]lim 1/(a-|w|) (1 - e^(-a))
= 0

Finally, parameterize γ₃ by γ₃(x) = a+ia - x, 0≤x≤2a. Then we have that:

[a→∞]lim |[γ₃]∫e^(iz)/(z-w) dz|
[a→∞]lim |[0, 2a]∫e^(i(a+ia - x))/(a + ia - x -w) (-1) dx|
≤ [a→∞]lim [0, 2a]∫|e^(i(a+ia - x))/(a + ia - x -w) (-1)| dx
= [a→∞]lim [0, 2a]∫e^(-a)/|a + ia - x - w| dx
≤ [a→∞]lim [0, 2a]∫e^(-a)/||a + ia - x| - |w|| dx
≤ [a→∞]lim [0, 2a]∫e^(-a)/(a-|w|) dx (since |a + ia - x| = ia + some real number, so a ≤ |a + ia - x|)
= [a→∞]lim 2ae^(-a)/(a-|w|)
= [a→∞]lim 2a/(a-|w|) e^(-a)
= 0 (since e^(-a) → 0 as a→∞)

So since the integral over γ₂ + γ₃ + γ₄ goes to 0 as a→∞, we have that [a→∞]lim [γ₁ + γ₂ + γ₃ + γ₄]∫e^(iz)/(z-w) dz = [a→∞]lim [γ₁]∫e^(iz)/(z-w) dz = [a→∞]lim [-a, a]∫e^(ix)/(x-w) dx. Thus by transitivity we have [a→∞]lim [-a, a]∫e^(ix)/(x-w) dx = {0 if Im(w) < 0, 2πie^(iw) if Im (w) > 0}

Now, to handle the other integral, namely [a→∞]lim [-a, a]∫e^(-iz)/(z-w) dz, you will again calculate the value of a related contour integral. Specifically, let γ₅ be the straight line from -a to a, γ₆ be the straight line from a to a - ai, γ₇ be the straight line from a - ai to -a - ai, and γ₈ be the straight line from -a - ai to -a. So the path γ₅ + γ₆ + γ₇ + γ₈ traces out a large rectangle in the _lower_ half-plane, and in the clockwise orientation (which is the opposite of the usual). Now, if Im (w) is positive, then w will never be on the inside of the rectangle, so e^(-iz)/(z-w) will be analytic on the interior [γ₅ + γ₆ + γ₇ + γ₈]∫e^(-iz)/(z-w) dz will be zero. Conversely, if Im(w) is negative, then for all sufficiently large values of a, w will be on the inside of the rectangle, so [γ₅ + γ₆ + γ₇ + γ₈]∫e^(-iz)/(z-w) dz = -2πi Res (e^(-iz)/(z-w), w) = -2πie^(-iw) (the negative sign comes from the fact that we are tracing the rectangle clockwise instead of counterclockwise). As with the first contour integral, we have that [γ₆ + γ₇ + γ₈]∫e^(-iz)/(z-w) dz approaches 0 as a →∞ (I'll leave the proof of this to you -- the methods are very similar to the methods used on the first integral -- namely, take the absolute value and keep making weaker and weaker estimates until you get something you can integrate). So we have:

[a→∞]lim [-a, a]∫e^(-iz)/(z-w) dz = [a→∞]lim [γ₅ + γ₆ + γ₇ + γ₈]∫e^(-iz)/(z-w) dz = {0 if Im (w) > 0, -2πie^(-iw) if Im(w)<0}. So now returning at last to the original integral:

1/2 [a→∞]lim [-a, a]∫e^(iz)/(z-w) dz + [-a, a]∫e^(-iz)/(z-w) dz
1/2 ({0 if Im (w) < 0, 2πie^(iw) if Im(w) > 0} + {0 if Im (w) > 0, -2πie^(-iw) if Im(w)<0})

{πie^(iw) if Im(w) > 0, -2πie^(-iw) if Im(w)<0}

And we are done.

Answer 2

On Turan Quadrature Formulas for the Chebyshev Weight*

Answer 3

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