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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!

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    Hint: Let $u = \log x$, then $du = \frac 1x\, dx$.2012-03-25
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    this integral is undefined .2012-03-25
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    OK but this leads to the expression minus infinity plus infinity and then what?2012-03-25
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    I want to clearly see why this integral diverges2012-03-25
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    Because (for example) $\int_e^\infty \frac 1x \, dx$ diverges and you integrand is larger.2012-03-25
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    @martini, but in the interval $(0,e)$, the integrand takes large negative values though. How do you account for that?2012-03-25
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    @FortuonPaendrag If we are not considering the PV, AFAIK such an integral diverges if it does "on one side".2012-03-25
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    @martini Ah, thank you. Like an alternating sum whose terms are only getting bigger in absolute value?.It makes sense now. I should have given it a little more thought.2012-03-25

5 Answers 5

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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).

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    Brilliant! I am convinced. Thank you, thank you, thank you :-)2012-03-25
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    @Halina, Glad to Help!2012-03-25
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    Just a moment: from your step 2, it follows that I=-infinity plus infinity so we are back to square number one again?2012-03-25
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    that is, $I=-\infty +\infty $ because $\int_{0}^{\infty }f\left( u\right) du=\infty $2012-03-25
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    @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 $\int_{0}^{\infty }f(u)du=\infty$.2012-03-25
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    @Halina $\int_{-\infty }^{+\infty}f(u)du$ is undefined.2012-03-25
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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 $0$.

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    How can it be both?2012-03-25
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    @Halina I said EITHER underfined, OR taking the PV $0$. Do some research on the Principal Value.2012-03-25
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    Ok but can you really apply the substitution rule to improper integrals?2012-03-25
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    @Halina, Yes, you can.2012-03-25
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    Thank you so very, very much :-)2012-03-25
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    Now, please look at this:2012-03-25
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    If $u=logx$, then $du=\frac{1}{x}dx$ so $I=\int_{0}^{\infty }\frac{\left( \log x\right) ^{40021}}{dx}=\int_{-\infty }^{\infty }u^{40021}du=\int_{-\infty }^{0}u^{40021}du+\int_{0}^{\infty }u^{40021}du=\lim_{t\rightarrow -\infty }\int_{t}^{0}u^{40021}du+\lim_{t\rightarrow \infty }\int_{0}^{t}u^{40021}du=\lim_{t\rightarrow -\infty }\left( -\left( \frac{t^{400022}}{400022}\right) \right) +\lim_{t\rightarrow \infty }\left( \frac{t^{400022}}{400022}\right) =-\infty +\infty $2012-03-25
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    so how can we conclude from the above that the integral I is divergent ?2012-03-25
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    I got it now> Thank you for your patience :-)2012-03-25
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    Always double check with what you see with other tools such as Wolfram Alpha, according to which the integral does not converge.2012-03-25
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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 $I_1 =0$

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    that looks like right solution...2014-02-05
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    This is deceptive. When we write $\int_{0}^{\infty}$ we are really doing $\lim_{b\to\infty} \int_{0}^b$ and so what we are really doing in this "proof" is adding two functions ($\int_{0}^b$ and $\int_{-b}^0$) whose limits don't exist and claiming that their sum exists.2014-06-27
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\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.

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    What about the other infinity? You have something wrong there. You get $\infty - \infty$, which is indeterminate, so you need to consider the PV.2012-03-25
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    exactly, this is what I meant, too2012-03-25
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    @Peter: there is no *need* to consider the Principal Value. It is *possible* to consider the the Principal Value. [Technically](http://en.wikipedia.org/wiki/Improper_integral#Convergence_of_the_integral), 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.2012-03-25
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    Actually, $\left[\frac{u^{40022}}{40022}\right]_{-\infty}^{s}=-\infty$, so $\lim\limits_{s\to\infty}\left[\frac{u^{40022}}{40022}\right]_{-\infty}^{s}=-\infty$2012-03-25
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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.

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    you might wanna look at my answer.2012-04-29