Question

Math

Posted 4 months ago

```
Consider the following integral:
$I=\int_{-\infty}^{\infty} \frac{\cos x}{\left(4 x^{2}-9 \pi^{2}\right)\left(x^{2}+9\right)} d x$
(a) Locate and characterize all singularities of the integrand in the complex $\mathrm{z}$ plane.
(b) Locate and characterize all singularities of the integrand when $\cos x$ is replaced by $e^{j x}$ (and $x$ by the complex variable $z$ ), i.e. $\frac{e^{j z}}{\left(4 z^{2}-9 \pi^{2}\right)\left(z^{2}+9\right)}$.
(c) Define the contours required to evaluate the given integral $\int_{-\infty}^{\infty} \frac{e^{j x}}{\left(4 x^{2}-9 \pi^{2}\right)\left(x^{2}+9\right)} d x$.
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Identify the integrand: $\frac{\cos x}{(4x^2 - 9\pi^2)(x^2 + 9)}$

step 2

Set the denominators to zero to find singularities: $4x^2 - 9\pi^2 = 0$ and $x^2 + 9 = 0$

step 3

Solve $4x^2 - 9\pi^2 = 0$: $x = \pm \frac{3\pi}{2}$

step 4

Solve $x^2 + 9 = 0$: $x = \pm 3i$

step 5

The singularities are $z = \pm \frac{3\pi}{2}$ (real) and $z = \pm 3i$ (imaginary)

```
(b) Locate and characterize all singularities of the integrand when $\cos x$ is replaced by $e^{jx}$ (and $x$ by the complex variable $z$).
```

step 1

Identify the new integrand: $\frac{e^{jz}}{(4z^2 - 9\pi^2)(z^2 + 9)}$

step 2

The singularities remain the same as in part (a): $z = \pm \frac{3\pi}{2}$ and $z = \pm 3i$

step 3

Characterize the singularities: They are simple poles

```
(c) Define the contours required to evaluate the given integral $\int_{-\infty}^{\infty} \frac{e^{jx}}{(4x^2 - 9\pi^2)(x^2 + 9)} dx$.
```

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 $z = 3i$ and $z = -3i$

step 4

Evaluate the residues at these poles to find the integral

Answer

The singularities of the integrand are $z = \pm \frac{3\pi}{2}$ and $z = \pm 3i$. When $\cos x$ is replaced by $e^{jx}$, 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 $z = 3i$ and $z = -3i$.

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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