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Chapter 4: Problem 1

Evaluate the following real integrals. (a) \(\int_{0}^{\infty} \frac{d x}{x^{2}+a^{2}}, \quad a^{2}>0\) (Verify your answer by using usual antiderivatives.) (b) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)^{2}}, \quad a^{2}>0\) (c) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}, \quad a^{2}, b^{2}>0\) (d) \(\int_{0}^{\infty} \frac{d x}{x^{6}+1}\)

Short Answer

Expert verified
Part (a): \(\frac{\text{蟺}}{2a}\)Part (b): \(\frac{\text{蟺}}{4a^3}\)Part (c): \(\frac{\text{蟺}}{(b^2 - a^2)} (\frac{1}{b} - \frac{1}{a})\)Part (d): \(\frac{\text{蟺}}{3}\)

Step by step solution

01

Integral Setup for Part (a)

Consider the integral \(\int_{0}^{\text{鈭瀩} \frac{dx}{x^2 + a^2}\). Identify the antiderivative for the integrand.
02

Antiderivative Calculation

The antiderivative of \(\frac{1}{x^2 + a^2}\) is \(\frac{1}{a} \tan^{-1} \frac{x}{a}\). We use this antiderivative to find the definite integral.
03

Evaluating Definite Integral for Part (a)

Now, evaluate \(\frac{1}{a} [ \tan^{-1} \frac{x}{a} ]_{0}^{\text{鈭瀩}\). This results in \(\frac{\pi}{2a} - 0\), which simplifies to \(\frac{\pi}{2a}\).
04

Integral Setup for Part (b)

Consider the integral \(\int_{0}^{\text{鈭瀩} \frac{dx}{(x^2 + a^2)^2}\). Use an appropriate trigonometric substitution or compare with known integrals.
05

Using Trigonometric Substitution for Part (b)

Let \(x = a \tan(\theta)\). By transforming the integrand and applying trigonometric identities, the integral simplifies using known integral results.
06

Evaluate Integral for Part (b)

Transform the limits and evaluate the resulting integral. The answer is \(\frac{\text{蟺}}{4a^3}\).
07

Integral Setup for Part (c)

Consider the integral \(\text{鈭珆_{0}^{\text{鈭瀩} \frac{dx}{(x^2 + a^2)(x^2 + b^2)}\). Use partial fraction decomposition to simplify the integrand.
08

Apply Partial Fractions for Part (c)

Express \( \frac{1}{(x^2 + a^2)(x^2 + b^2)} \) as \( \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2} \) and solve for A and B.
09

Evaluate Each Integral for Part (c)

Evaluate each part separately. Using previous results, the solution simplifies to \(\frac{\text{蟺}}{(b^2 - a^2)} (\frac{1}{b} - \frac{1}{a})\).
10

Integral Setup for Part (d)

Consider the integral \(\text{鈭珆_{0}^{\text{鈭瀩} \frac{dx}{x^6 + 1}\). Identify a strategy, such as contour integration or residue theorem.
11

Applying Contour Integration for Part (d)

Map the integral to a contour in the complex plane. Identify the residues at the poles within the contour and calculate the integral.
12

Evaluate Residues for Part (d)

Sum the residues and multiply by \(2 \text{蟺i}\). This provides the result \(\frac{\text{蟺}}{3 \text{sin}(\frac{\text{蟺}}{6})} = \frac{\text{蟺}}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Contour Integration
Contour integration is a technique used to evaluate complex integrals by integrating along a contour in the complex plane. This method is particularly useful when dealing with functions that have poles.

The integral \(\text{\int}_{0}^{\text{鈭瀩} \frac{dx}{x^6 + 1}\) can be tackled via contour integration. We map the integral to a contour in the complex plane and identify the residues of the integrand at its poles. Summing these residues and multiplying by \(2 \pi i\) gives us the final result: \(\frac{\pi}{3}\). This approach can significantly simplify real integrals, making them more manageable.

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Most popular questions from this chapter

Establish the following relationships, where we use the notation \(\mathcal{L}(f(x))\) \(\equiv \hat{F}(s):\) (a) $$ \mathcal{L}\left(e^{a x} f(x)\right)=\hat{F}(s-a) \quad a>0 $$ (b) \(\quad \mathcal{L}(f(x-a) H(x-a))=e^{-a s} \hat{F}(s) \quad a>0\), where $$ H(x)= \begin{cases}1 & x \geq 0 \\ 0 & x<0\end{cases} $$ (c) Use the convolution product formula for Laplace transforms to show that the inverse Laplace transform of \(\hat{H}(s)=1 /\left(s^{2}\left(s^{2}+\omega^{2}\right)\right), \omega>0\) satisfies $$ \begin{aligned} h(x) &=\frac{1}{\omega} \int_{0}^{x} x^{\prime} \sin \omega\left(x-x^{\prime}\right) d x^{\prime} \\ &=\frac{x}{\omega^{2}}-\frac{\sin \omega x}{\omega^{3}} \end{aligned} $$ Verify this result by using partial fractions.

Obtain the Fourier transform of the following functions, and thereby show that the Fourier transforms of hyperbolic secant functions are also related to hyperbolic secant functions. (a) sech \(\left[a\left(x-x_{0}\right)\right] e^{i \omega x}, \quad a, x_{0}, \omega\) real (b) \(\operatorname{sech}^{2}\left[a\left(x-x_{0}\right)\right], \quad a, x_{0}\) real

Given the ODE $$ z y^{\prime \prime}+(2 r+1) y^{\prime}+z y=0 $$ look for a contour representation of the form \(y=\int_{C} e^{z \zeta} v(\zeta) d \zeta\). (a) Show that if \(C\) is a closed contour and \(v(\zeta)\) is single valued on this contour, then it follows that \(v(\zeta)=A\left(\zeta^{2}+1\right)^{r-1 / 2}\) (b) Show that if \(y=z^{-s} w\), then when \(s=r, w\) satisfies Bessel's equation \(z^{2} w^{\prime \prime}+z w^{\prime}+\left(z^{2}-r^{2}\right) w=0\), and a contour representation of the solution is given by $$ w=A z^{r} \oint_{C} e^{z \zeta}\left(\zeta^{2}+1\right)^{r-1 / 2} d \zeta $$ Note that if \(r=n+1 / 2\) for integer \(n\), then this representation yields the trivial solution. We take the branch cut to be inside the circle \(C\) when \((r-1 / 2) \neq\) integer.

Show $$ \int_{0}^{2 \pi} \cos ^{2 n} \theta d \theta=\int_{0}^{2 \pi} \sin ^{2 n} \theta d \theta=\frac{2 \pi}{2^{2 n}} B_{n}, \quad n=1,2,3, \ldots $$ where \(B_{n}=2^{2 n}(1 \cdot 3 \cdot 5 \cdots(2 n-1)) /(2 \cdot 4 \cdot 6 \cdots(2 n))\). (Hint: the fact that in the binomial expansion of \((1+w)^{2 n}\) the coefficient of the term \(w^{n}\) is \(B_{n}\).)

In this problem we obtain the Green's function of Laplace's equation in the upper half plane, \(-\infty
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