Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

Please show me the detailed solution to the question:

Compute the value of $$\int_{0}^{\infty }\frac{\left( \ln x\right) ^{40021}}{x}dx$$

Thank you a million!

share|improve this question
2  
Hint: Let $u = \log x$, then $du = \frac 1x\, dx$. – martini Mar 25 '12 at 7:57
1  
this integral is undefined . – pedja Mar 25 '12 at 8:03
OK but this leads to the expression minus infinity plus infinity and then what? – Halina Mar 25 '12 at 8:04
I want to clearly see why this integral diverges – Halina Mar 25 '12 at 8:05
Because (for example) $\int_e^\infty \frac 1x \, dx$ diverges and you integrand is larger. – martini Mar 25 '12 at 8:11
show 3 more comments

5 Answers

up vote 2 down vote

Edited. Let $u=\log x$ (as per martini's hint) and $f(u)=u^{40021}$. Then

  1. $$\begin{equation*} I:=\int_{0}^{\infty }\frac{\left( \log x\right) ^{40021}}{x}dx=\int_{-\infty}^{\infty}f(u)du, \end{equation*}$$
  2. Function $f$ is odd, $f(-u)=-f(u)$. These integrals don't converge $$ \begin{equation*} \int_{-\infty }^{0}f(u)du=-\int_{0}^{\infty }f(u)du. \end{equation*}$$

The integral $I$ is undefined (as commented by pedja).

share|improve this answer
Brilliant! I am convinced. Thank you, thank you, thank you :-) – Halina Mar 25 '12 at 10:44
@Halina, Glad to Help! – Américo Tavares Mar 25 '12 at 10:47
Just a moment: from your step 2, it follows that I=-infinity plus infinity so we are back to square number one again? – Halina Mar 25 '12 at 11:22
that is, $I=-\infty +\infty $ because $\int_{0}^{\infty }f\left( u\right) du=\infty $ – Halina Mar 25 '12 at 11:25
@Halina Right. Both $\int_{-\infty }^{0}f(u)du$ and $\int_{0}^{\infty }f(u)du$ diverge. $\int_{-\infty }^{0}f(u)du=-\infty$ and $\int_{0}^{\infty }f(u)du=\infty$. – Américo Tavares Mar 25 '12 at 12:08
show 1 more comment
up vote 1 down vote

Make the change of variables suggested. You'll end up with $$\int\limits_{-\infty}^\infty u^{40021}du$$ Then the integral is either undefined or taking the principal value, it is $0$.

share|improve this answer
How can it be both? – Halina Mar 25 '12 at 8:42
@Halina I said EITHER underfined, OR taking the PV $0$. Do some research on the Principal Value. – Peter Tamaroff Mar 25 '12 at 8:54
Ok but can you really apply the substitution rule to improper integrals? – Halina Mar 25 '12 at 10:00
@Halina, Yes, you can. – Américo Tavares Mar 25 '12 at 10:27
Thank you so very, very much :-) – Halina Mar 25 '12 at 10:45
show 5 more comments
up vote 0 down vote

\begin{align*} \int_{0}^{\infty}\frac{{\log x}^{40021}}{x}dx &=\lim_{s\to\infty}\int_{0}^{s}\frac{{\log x}^{40021}}{x}dx\\ &=\lim_{s\to\infty}\int_{-\infty}^{s}u^{40021}du\\ &=\lim_{s\to\infty}\left[\frac{u^{40022}}{40022}\right]_{-\infty}^{s}\\ &=\infty. \end{align*} In this i take $\log x= u$. And this example is a improper integral of the third kind.

share|improve this answer
What about the other infinity? You have something wrong there. You get $\infty - \infty$, which is indeterminate, so you need to consider the PV. – Peter Tamaroff Mar 25 '12 at 8:30
exactly, this is what I meant, too – Halina Mar 25 '12 at 8:44
3  
@Peter: there is no need to consider the Principal Value. It is possible to consider the the Principal Value. Technically, the integral diverges: $\int_0^\infty\frac{\log(x)^{40021}}{x}\mathrm{d}x=\int_0^1\frac{\log(x)^{40021}‌​}{x}\mathrm{d}x+\int_1^\infty\frac{\log(x)^{40021}}{x}\mathrm{d}x$ and both of these integrals diverge. – robjohn♦ Mar 25 '12 at 11:02
Actually, $\left[\frac{u^{40022}}{40022}\right]_{-\infty}^{s}=-\infty$, so $\lim\limits_{s\to\infty}\left[\frac{u^{40022}}{40022}\right]_{-\infty}^{s}=-\in‌​fty$ – robjohn♦ Mar 25 '12 at 11:09
up vote 0 down vote

Let me introduce you to elegant mathematics!

$$I_1=\int\limits_{0}^{\infty }\frac{\left( \ln x\right) ^{40021}}{x}dx$$

put $x=\frac{1}{t}$, to get

$$\int\limits_{\infty}^{0 }\frac{\left( \ln \frac{1}{t}\right) ^{40021}}{\frac{1}{t}}.\frac{-dt}{t^2}$$

$$\int\limits_{\infty}^{0 }\frac{\left( \ln t\right) ^{40021}}{t}dt=-\int\limits_{0}^{\infty }\frac{\left( \ln t\right) ^{40021}}{t}dt=I_1$$

so,

$I_1=-\int\limits_{0}^{\infty }\frac{\left( \ln t\right) ^{40021}}{t}dt=-\int\limits_{0}^{\infty }\frac{\left( \ln x\right) ^{40021}}{x}dx$

and

$I_1=\int\limits_{0}^{\infty }\frac{\left( \ln x\right) ^{40021}}{x}dx$

addint the 2 gives $2I_1=0$, hence $I_1 =0$

share|improve this answer
up vote 0 down vote

Since this is an exercise on improper integrals, it is natural to replace the upper and lower limits by $R$, $\frac{1}{R}$ respectively and define the integral to be the limit as $R \rightarrow \infty$ . Then write the integral as the sum of the integral from $\frac{1}{R}$ to $1$ and from $1$ to $R$. In the second integral make the usual transformation replacing $x$ by $\frac{1}{x}$. The two integrals cancel.

share|improve this answer
you might wanna look at my answer. – Tomarinator Apr 29 '12 at 15:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.