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Question
Math
Posted 4 months ago

Consider the following integral:
I=cosx(4x29π2)(x2+9)dx
I=\int_{-\infty}^{\infty} \frac{\cos x}{\left(4 x^{2}-9 \pi^{2}\right)\left(x^{2}+9\right)} d x

(a) Locate and characterize all singularities of the integrand in the complex z\mathrm{z} plane.
(b) Locate and characterize all singularities of the integrand when cosx\cos x is replaced by ejxe^{j x} (and xx by the complex variable zz ), i.e. ejz(4z29π2)(z2+9)\frac{e^{j z}}{\left(4 z^{2}-9 \pi^{2}\right)\left(z^{2}+9\right)}.
(c) Define the contours required to evaluate the given integral ejx(4x29π2)(x2+9)dx\int_{-\infty}^{\infty} \frac{e^{j x}}{\left(4 x^{2}-9 \pi^{2}\right)\left(x^{2}+9\right)} d x.
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Answer from Sia
Posted 4 months ago
Solution by Steps
step 1
Identify the integrand: cosx(4x29π2)(x2+9) \frac{\cos x}{(4x^2 - 9\pi^2)(x^2 + 9)}
step 2
Set the denominators to zero to find singularities: 4x29π2=0 4x^2 - 9\pi^2 = 0 and x2+9=0 x^2 + 9 = 0
step 3
Solve 4x29π2=0 4x^2 - 9\pi^2 = 0 : x=±3π2 x = \pm \frac{3\pi}{2}
step 4
Solve x2+9=0 x^2 + 9 = 0 : x=±3i x = \pm 3i
step 5
The singularities are z=±3π2 z = \pm \frac{3\pi}{2} (real) and z=±3i z = \pm 3i (imaginary)


 (b) Locate and characterize all singularities of the integrand when cosx\cos x is replaced by ejxe^{jx} (and xx by the complex variable zz).
step 1
Identify the new integrand: ejz(4z29π2)(z2+9) \frac{e^{jz}}{(4z^2 - 9\pi^2)(z^2 + 9)}
step 2
The singularities remain the same as in part (a): z=±3π2 z = \pm \frac{3\pi}{2} and z=±3i z = \pm 3i
step 3
Characterize the singularities: They are simple poles


 (c) Define the contours required to evaluate the given integral ejx(4x29π2)(x2+9)dx \int_{-\infty}^{\infty} \frac{e^{jx}}{(4x^2 - 9\pi^2)(x^2 + 9)} dx .
step 1
Use the residue theorem to evaluate the integral
step 2
Choose a semicircular contour in the upper half-plane to avoid the poles on the real axis
step 3
The contour should enclose the poles at z=3i z = 3i and z=3i z = -3i
step 4
Evaluate the residues at these poles to find the integral
Answer
The singularities of the integrand are z=±3π2 z = \pm \frac{3\pi}{2} and z=±3i z = \pm 3i . When cosx\cos x is replaced by ejxe^{jx}, the singularities remain the same and are simple poles. To evaluate the integral, use a semicircular contour in the upper half-plane that encloses the poles at z=3i z = 3i and z=3i z = -3i .
Key Concept
Singularities and Contour Integration
Explanation
The key concept involves identifying singularities of the integrand and using contour integration to evaluate the integral. The singularities are found by setting the denominator to zero, and the contour is chosen to apply the residue theorem effectively.

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