Integration--Finding Principal Value

[clickbb08]

I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.

 

[Dustinsfl]

I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.

First it should be
$$
\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = \frac{\pi}{3e^3}
$$

Let $f(z) = \dfrac{e^{iz}}{(z-3i)(z+3i)}$

Then
$$
\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = 2\pi i\sum_{\text{UHP}}\text{Res}_{z=z_j}f(z) + \pi i\sum_{\text{real axis}}\text{Res}_{z=z_j}f(z)
$$

UHP = upper half plane

 

[math771]

Hi there,

To solve this problem, we can use the Residue Theorem. First, let's rewrite the integral as a complex contour integral:

∫Cosx/(x^2+9)dx = ∫Cosz/(z^2+9)dz

Where z = x+iy and dz = dx + idy.

Next, we can define a semicircular contour C with radius R, centered at the origin, and with endpoints at -R and R. This contour encloses the poles at z = 3i and z = -3i.

Using the Residue Theorem, we can write the integral as the sum of the residues at each pole:

∫Cosz/(z^2+9)dz = 2πi(Res(z=3i) + Res(z=-3i))

To calculate the residues, we can use the formula:

Res(z=z0) = lim(z→z0) (z-z0)f(z)

Applying this formula to our integral, we get:

Res(z=3i) = lim(z→3i) (z-3i)Cosz/(z^2+9)

= (3i-3i)Cos(3i)/(9+9)

= -iCos(3i)/6

Similarly, Res(z=-3i) = iCos(-3i)/6 = iCos(3i)/6

Substituting these values into our original integral, we get:

∫Cosx/(x^2+9)dx = 2πi(-iCos(3i)/6 + iCos(3i)/6)

= π/3e^2

Therefore, the P.V. of the integral is indeed equal to (π/3e^2). I hope this helps! Let me know if you have any further questions.