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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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    @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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    @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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    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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    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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    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