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Chapter 6: Problem 20

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} e^{-a x} \sin b x d x, \quad a>0 $$

Short Answer

Expert verified
The result of the integral is zero, hence the integral is convergent and its value is \(0\).

Step by step solution

01

Establishment of the Integral

Firstly, re-write the integral to recognise that you are dealing with the product of two functions, \(e^{-a x}\) and \(\sin b x\). After recognizing this, assign \(u=e^{-a x}\) and \(v=\sin b x\). Consequently, using the formula for integral by parts, \(\int u dv = uv - \int v du\), the integral is represented as: \[ \int_{0}^{\infty} e^{-a x} \sin b x d x = [-e^{-ax} \sin bx]_{0}^{\infty} - \int_{0}^{\infty} -e^{-ax} b \cos bx dx \]
02

Performing the Operation

Here we perform the operation of the integral, by substituting \(x = 0\) and \(x = \infty\) into \(-e^{-ax} \sin bx\). After substituting and simplifying, it results to \(0\). So, the left-over part of the calculation is \(-b \int_{0}^{\infty} e^{-ax} \cos bx dx\). We can apply the same integral by parts method once again to this new integral.
03

Second Integration by Parts

Assign \(u=e^{-a x}\) and \(v=\cos b x\). The new integral, \(-b \int_{0}^{\infty} e^{-a x} \cos bx dx\), is represented as: \[-b [(e^{-ax} \cos bx / a]_{0}^{\infty} + b/a \int_{0}^{\infty} e^{-ax} \sin bx dx) \] The first term here also goes to \(0\) when we substitute \(x = 0\) and \(x = \infty\). What remains is the improper integral we started with, so you can equate this to the original improper integral, and solve for it.
04

Solve Equation

Set a variable, \(I\), to represent the original improper integral. Therefore, \(I = -b [(e^{-ax} \cos bx / a]_{0}^{\infty} + b/a I). Re-arrange this equation to solve for \(I\), thus: \(I (1+ (b^2/a)) = 0. Solving this equation results in the final answer: \(I = 0 /(1+ b^2/a) \) or simply, \(I = 0\).

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Most popular questions from this chapter

(a) Let \(f^{\prime}(x)\) be continuous. Show that \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)\) (b) Explain the result of part (a) graphically.

Rewrite the improper integral as a proper integral using the given \(u\) -substitution. Then use the Trapezoidal Rule with \(n=5\) to approximate the integral. $$ \int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x, \quad u=\sqrt{x} $$

Evaluate the definite integral. $$ \int_{0}^{\pi / 4} \tan ^{3} x d x $$

Define the terms converges and diverges when working with improper integrals.

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=\cosh a t $$
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