How to evaluate one improper integral 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!
 A: Edited. 
Let $u=\log x$ (as per martini's hint) and $f(u)=u^{40021}$. Then 


*

*$$\begin{equation*}
I:=\int_{0}^{\infty }\frac{\left( \log x\right) ^{40021}}{x}dx=\int_{-\infty}^{\infty}f(u)du,
\end{equation*}$$

*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).
A: 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 $0$.
A: 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 $I_1 =0$
A: \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.
A: 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. 
