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Chapter 7: Problem 4

For what values of \(p\) does \(\int_{1}^{\infty} x^{-p} d x\) converge?

Short Answer

Expert verified
Answer: The integral converges for values of \(p\) such that \(p<1\).

Step by step solution

01

Set up the improper integral

To determine the convergence of the improper integral \(\int_{1}^{\infty} x^{-p} dx\), first we rewrite the integral with limits for x approaching infinity: \[\int_{1}^{\infty} x^{-p} dx = \lim_{b \to \infty} \int_{1}^{b} x^{-p} dx\]
02

Calculate the antiderivative of the given function

Next, we find the antiderivative of \(x^{-p}\). This can be accomplished using the power rule, which states that the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}+C\), when \(n\neq -1\). Since we want to find the antiderivative of \(x^{-p}\), we find: \[ \int x^{-p} dx = \frac{x^{-p+1}}{-p+1}+C = \frac{x^{1-p}}{1-p}+C\]
03

Calculate the limit of the antiderivative as b approaches infinity

Now we evaluate the antiderivative at the limits of integration and take the limit as b approaches infinity: \[\lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_{1}^{b} = \lim_{b \to \infty} \left(\frac{b^{1-p}}{1-p}-\frac{1^{1-p}}{1-p}\right)\]
04

Determine the values of p for which the limit converges

The limit of the antiderivative will converge if the first term \(b^{1-p}\) approaches zero as \(b \to \infty\). The exponent of b, \(1-p\), determines whether this term approaches zero or infinity. If \(1-p>0\), then \(p<1\) and \(b^{1-p} \to 0\) as \(b \to \infty\). In this case, the limit converges. If \(1-p<0\), then \(p>1\) and \(b^{1-p} \to \infty\) as \(b \to \infty\). In this case, the limit diverges. If \(1-p=0\), then \(p=1\) and \(b^{1-p} = 1\). The limit will be undefined in this case, due to the term \(\frac{x^{1-p}}{1-p}\) having a denominator of zero. So, the integral diverges for \(p=1\).
05

Conclusion

So, the integral \(\int_{1}^{\infty} x^{-p} dx\) converges for values of \(p\) such that \(p<1\) and diverges for values of \(p\) such that \(p \ge 1\).

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Most popular questions from this chapter

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

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