Integration--Finding Principal Value

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SUMMARY

The principal value (P.V.) of the integral from negative infinity to infinity of the function Cos(x)/(x^2 + 9) is definitively calculated as (π/3e^3). The integral involves identifying isolated singular points (ISP) at -3i and 3i. The function f(z) = e^(iz)/((z-3i)(z+3i)) is utilized to evaluate the integral using residue calculus, specifically focusing on contributions from the upper half plane (UHP) and the real axis.

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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.
 
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clickbb08 said:
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
 
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