Does the integral converge? $\int_1^\infty \frac{\ln(1+x)}{x^2}dx$ Does the integral converge? $$\int_1^\infty \frac{\ln(1+x)}{x^2}dx$$
Well, I used integration by parts and got to $\ln4$, which means it is clearly converges. but I want to try another approach as this integral is positive for every $x \in [1,\infty]$.
I wanted to ask you if it is possible to change $\ln(1+x)$ with $\ln(e)$ if I say that $1 + x = e$ , Thus $ x = e - 1 $ and then the integral is:
$$\int_{e-1}^\infty \frac{1}{x^2}dx$$ And then I can say that because the integration limit is upto infinity and because that function is closed somehow by $m < f(x) < M$ then, the second integral converges, and thus the first one converges too.
What do you think?
 A: We can show that the integral converges without our evaluating it.
Note that we can write $\log(1+x)$ as 
$$\log(1+x)=\log(x)+\log\left(1+\frac1x\right) \tag 1$$
Now, recall that the logarithm function satisfies the inequality
$$\log(x)\le x-1<x \tag 2$$   
for $x>0$.  
Therefore, for any number $\alpha>0$, we have
$$\log(x^\alpha)<x^\alpha\implies \log(x)<\frac{x^\alpha}{\alpha} \tag 3$$ 
We can choose any positive $\alpha$.  Let's choose arbitrarily $\alpha <1/2$.  
Then, using $(1)$, $(2)$ and $(3)$ with $\alpha =1/2$, we obtain
$$\log(x+1)<2x^{1/2}+\frac1x$$
Finally, 
$$\frac{\log(1+x)}{x^2}<2x^{-3/2}+\frac1{x^3}$$
Since the integral $$\int_1^\infty \left(2x^{-3/2}+\frac1{x^3}\right)\,dx$$
exists, then the integral $$\int_1^\infty \frac{\log(1+x)}{x^2}\,dx$$
exists
A: $$\int_{1}^{\infty}\frac{\ln(1+x)}{x^2}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{1}^{n}\frac{\ln(1+x)}{x^2}\space\text{d}x=$$

Integrate by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where:
$$f=\ln(1+x),\text{d}g=\frac{1}{x^2}\space\text{d}x,\text{d}f=\frac{1}{1+x}\space\text{d}x,g=-\frac{1}{x}$$

$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}+\int_{1}^{n}\frac{1}{x+x^2}\space\text{d}x\right)=$$
$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}+\int_{1}^{n}\frac{1}{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}\space\text{d}x\right)=$$

Substitute $u=x+\frac{1}{2}$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=1+\frac{1}{2}=\frac{3}{2}$ and upper bound $u=n+\frac{1}{2}$:

$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}+\int_{\frac{3}{2}}^{n+\frac{1}{2}}\frac{1}{u^2-\frac{1}{4}}\space\text{d}u\right)=$$
$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}+4\int_{\frac{3}{2}}^{n+\frac{1}{2}}\frac{1}{1-4u^2}\space\text{d}u\right)=$$

Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.
This gives a new lower bound $s=2\cdot\frac{3}{2}=3$ and upper bound $s=2n+1$:

$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}-2\int_{3}^{2n+1}\frac{1}{1-s^2}\space\text{d}s\right)=$$
$$\lim_{n\to\infty}\left(\left[-\frac{\ln(1+x)}{x}\right]_{1}^{n}-2\left[\text{arctanh}(s)\right]_{3}^{2n+1}\right)=$$
$$\lim_{n\to\infty}\left(\ln\left(\frac{4n}{1+n}\right)-\frac{\ln(1+n)}{n}\right)=\ln(4)$$
A: The integral trivially converges, since $0 < \ln(1+x) < \sqrt{x}$, for $x \geq 1$. Hence, we have
$$0 < \int_1^{\infty} \dfrac{\ln(1+x)}{x^2}dx < \int_1^{\infty} \dfrac{\sqrt{x}}{x^2}dx = \int_1^{\infty} x^{-3/2}dx = 2$$
We can in fact evaluate the integral. Let
\begin{align}
I(a) & = \int_1^{\infty} \dfrac{\ln(1+ax)}{x^2}dx\\
I'(a) & = \lim_{R \to \infty}\int_1^R \dfrac{dx}{x(1+ax)} = \lim_{R \to \infty} \left(\int_1^R \dfrac{dx}x - \int_1^R \dfrac{adx}{1+ax}\right) = \lim_{R \to \infty}\left(\ln(R) - \ln(1+aR) + \ln(1+a) \right)\\
& =\lim_{R \to \infty}\left( -\ln\left(\dfrac1R+a\right) + \ln(1+a) \right) = \ln(a+1) - \ln(a) \,\,\, (\spadesuit)
\end{align}
Note that $I(0) = 0$. Now integrating $(\spadesuit)$ from $0$ to $1$, we have
\begin{align}
I(1) - I(0) & = \int_0^1 \ln(1+a)da - \int_0^1 \ln(a)da = \left(\ln(4)-1\right) - \left(-1\right) = \ln(4)
\end{align}
Hence, we obtain that
$$I(1) = \int_1^{\infty} \dfrac{\ln(1+x)}{x^2}dx = \ln(4)$$
