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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 $du = \frac 1x\, dx$. – martini Mar 25 '12 at 7:57
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

5 Answers 5

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 :-) – 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

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? – Halina Mar 25 '12 at 8:42
@Halina I said EITHER underfined, OR taking the PV $0$. Do some research on the Principal Value. – Pedro 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

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

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

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


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


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

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

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that looks like right solution... – lowtech Feb 5 '14 at 4:12
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. – Alqatrkapa Jun 27 '14 at 5:26

\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. – Pedro Tamaroff Mar 25 '12 at 8:30
exactly, this is what I meant, too – Halina Mar 25 '12 at 8:44
@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

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. – Tomarinator Apr 29 '12 at 15:16

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