To calculate an improper integral Is $$\int_1^\infty \frac{\log x}{x^2}dx$$
finite? How to solve this?
 A: We integrate by parts
\begin{align*}
\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx &= \lim\limits_{t\to\infty}\int_{1}^{t}\frac{\ln(x)}{x^2}dx\\
&= \lim\limits_{t\to\infty}\left(\left[-\frac{\ln(x)}{x}\right]_{1}^{t}-\int_{1}^{t}-\frac{1}{x}\cdot\frac{1}{x}\text{ }dx\right)\\
&= \lim\limits_{t\to\infty}\left(\frac{\ln(1)}{1}-\frac{\ln(t)}{t}-\int_{1}^{t}-\frac{1}{x^2}dx\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left[\frac{1}{x}\right]_{1}^{t}\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left(\frac{1}{t}-\frac{1}{1}\right)\right)\\
&=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)
\end{align*}
As $t\to\infty$, $\frac{\ln(t)}{t}$ and $\frac{1}{t}$ both tend to $0$, so
$$\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)=1$$
A: If you don't like the $\log$, then you can turn it back to exponent by setting $u = \log x \implies x = e^u\implies dx = e^udu\implies I = \displaystyle \int_{0}^\infty ue^{-2u}(e^{u}du)= \displaystyle \int_{0}^\infty ue^{-u}du= -\displaystyle \int_{0}^\infty ud(e^{-u})=- \left(ue^{-u}|_{u=0}^\infty-\displaystyle \int_{0}^\infty e^{-u}du\right)=\displaystyle \int_{0}^\infty e^{-u}du = 1.$
A: Let $t= \ln x.$ When we differentiate this we get $dt=\frac{1}{x}dx$. Also $e^{t} = x.\Rightarrow \color{red}A$
By $\color{red}A$ we get $e^{t}dt=dx\Rightarrow \color{red}B$.
So by substituting $\color{red}B$ we get,
$$\int_{0}^{\infty} te^{-t}dt=\int_{0 }^{\infty} t\frac{d\left [ -e^{-t} \right ]}{dt}dt$$
By integration by parts we get,
$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{0}^{\infty}-\int_{0}^{\infty}-e^{-t}dt=\left [ -te^{-t} \right ]_{0 }^{\infty}+\int_{0}^{\infty}e^{-t}dt$$
$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{1}^{\infty}+\left [ -e^{-t} \right ]_{0}^{\infty}=\left [ -te^{-t}-e^{-t} \right ]_{0}^{\infty}={\infty}=\left [ -e^{-t}(t+1) \right ]_{0}^{\infty}$$
So we have to find $\lim_{t\to\infty}e^{-t}(t+1)$ value,
$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}$$
So by L'hopital rule,
$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}=\lim_{t\to\infty}\frac{1}{e^{t}}=0$$
$$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=\lim_{t \to \infty} \left [ -e^{-t}(t+1) \right ]_{0}^{t}$$
So we get $$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=0+1=1$$
So as our final answer we get,
$$\int_1^\infty \frac{\log x}{x^2}dx=1$$
I got this as my answer. If there is any miscalculations. Please correct me.
A: Use integration by parts. Keep trying different parts until it works.

*

*This doesn't work:

$u= \frac1x$, $dv=\frac{\ln(x)dx}{x}$
$du= -\frac{dx}{x^2}$, $v=\frac{\ln^2(x)}{2}$


*This doesn't work:

$u= \frac1{x^2}$, $dv=\frac{\ln(x)}{1}$
$du= (...)$, $v=(...)$


*This works

$u=\frac{\ln(x)}{1}$, $dv= \frac1{x^2}$
$du=\frac{1}{x}$, $v= -\frac1{x}$
