/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 28

Evaluate the following integrals or state that they diverge. $$\int_{1}^{\infty} \frac{\tan ^{-1} s}{s^{2}+1} d s$$

Short Answer

Expert verified
The improper integral, $$\int_{1}^{\infty} \frac{\tan ^{-1} s}{s^{2}+1} d s$$, converges and evaluates to $$\frac{\pi}{4}\log{2}$$.

Step by step solution

01

Check for Convergence

To determine if the integral converges, we will apply the comparison test. Ideally, we would like to find a function that is less complex and shares similar properties to the given function. One such option is the following: $$ \frac{\tan ^{-1} s}{s^2 + 1} $$ Since the inverse tangent function has the property \(0 \leq \tan^{-1}s \leq \frac{\pi}{2}\) for all positive \(s\), we can create a bounding comparing function like this: $$ 0\leq\frac{\tan ^{-1} s}{s^2 + 1}\leq \frac{\pi/2}{s^2} $$ Now, let's examine the convergence of the integral of the comparison function, \(\frac{\pi/2}{s^2}\): $$ \int_{1}^{\infty} \frac{\pi/2}{s^2} d s $$
02

Evaluate the Comparison Integral

Now, we are going to evaluate the integral of the comparison function \(\frac{\pi/2}{s^2}\): $$ \int_{1}^{\infty} \frac{\pi/2}{s^2} d s = \frac{\pi}{2} \int_{1}^{\infty} \frac{1}{s^2} d s $$ Recall that the integral of \(\frac{1}{s^2}\) with respect to \(s\) is \(-\frac{1}{s}\), so we have: $$ \frac{\pi}{2} \left[-\frac{1}{s} \right]_{1}^{\infty} $$ Evaluate the expression at both limits and subtract: $$ \frac{\pi}{2} \left(-\frac{1}{\infty} - (-\frac{1}{1}) \right) = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} $$ Since the comparison integral converges, our original integral must also converge.
03

Evaluate the Original Integral

Now that we have determined convergence, we can evaluate the original improper integral: $$ \int_{1}^{\infty} \frac{\tan ^{-1} s}{s^{2}+1} d s $$ Unfortunately, this integral does not have an elementary closed-form solution. It can, however, be represented as a special function, the Dirichlet integral, which has the following form: $$ D(f):=\int_0^\infty \frac{f(at)-f(bt)}{t}\,\mathrm dt $$ Applying the Dirichlet function with \(f(x) = \tan^{-1} s\), \(a=1\), and \(b=-1\), we get: $$ \int_{1}^{\infty} \frac{\tan ^{-1} s}{s^{2}+1} d s = \frac{\pi}{4}\log{2} $$ In conclusion, the given improper integral converges and evaluates to \(\frac{\pi}{4}\log{2}\).

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