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Find the values of p for which the integral converges. (Enter your answer as an inequality.) ...

Question:

a. Find the values of p for which the integral converges. (Enter your answer as an inequality.)

{eq}\int_e^{\infty} \frac {41}{x (\ln \ x)^p} dx {/eq}

b. Evaluate the integral for those values of {eq}p {/eq}.

Integration by Substitution

Integration by substitution is the way to solve the integral function by replacing a group of variables with another variable after integration again replacement of the original set of variables with assumed one.

Answer and Explanation:


Given data

  • The integral function is: {eq}\int\limits_e^\infty {\dfrac{{41}}{{x{{\left( {\ln x} \right)}^p}}}dx} {/eq}


(a)

Rewrite the integral function,

{eq}\begin{align*} I &= \int\limits_e^\infty {\dfrac{{41}}{{x{{\left( {\ln x} \right)}^p}}}dx} \\ &= \int\limits_e^\infty {\dfrac{{41}}{{{{\left( {\ln x} \right)}^p}}}\left( {\dfrac{{dx}}{x}} \right)} \end{align*} {/eq}


Let us assume that {eq}\ln x = t {/eq}.


Differentiate the above equation,

{eq}\begin{align*} \dfrac{1}{x} &= \dfrac{{dt}}{{dx}}\\ \dfrac{{dx}}{x} &= dt \end{align*} {/eq}


From the original integral,

{eq}\begin{align*} I &= \int\limits_1^\infty {\dfrac{{41}}{{{t^p}}}dt} \\ &= 41\int\limits_1^\infty {\dfrac{{dt}}{{{t^p}}}} \\ &= 41\left( {\dfrac{{{t^{ - p + 1}}}}{{ - p + 1}}} \right)_1^\infty \\ &= \dfrac{{41}}{{1 - p}}\left( {{t^{ - p + 1}}} \right)_1^\infty \end{align*} {/eq}


The value of above integral when {eq}p > 1 {/eq} is,

{eq}\begin{align*} I &= \dfrac{{41}}{{1 - p}}\left( {0 - 1} \right)\\ &= \dfrac{{41}}{{p - 1}} \end{align*} {/eq}


The value of above integral when {eq}p < 1 {/eq} is,

{eq}\begin{align*} I &= \dfrac{{41}}{{1 - p}}\left( \infty \right)\\ &= \infty \end{align*} {/eq}


Thus, the integral converges when {eq}p > 1 {/eq}.


(b)

Thus, the value of integral for those values of p is {eq}\dfrac{{41}}{{p - 1}} {/eq}.


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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