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!
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!
Edited. Let $u=\log x$ (as per martini's hint) and $f(u)=u^{40021}$. Then
The integral $I$ is undefined (as commented by pedja).
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 $0$.
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 $I_1 =0$
\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.
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.