MHB How Do We Determine the Convergence of These Complex Integrals?

[mathmari]

Hey!

I want to check the convergence of the following integrals:

1. $\displaystyle{\int_2^{\infty}\frac{1}{x\left (\log (x)\right )^2}dx}$
   
   We have that: \begin{equation*}\int_2^{\infty}\frac{1}{x\left (\log (x)\right )^2}dx=\lim_{b\rightarrow \infty}\int_2^b\frac{1}{x\left (\log (x)\right )^2}dx\end{equation*}
   
   To calculate the integral $\int_2^b\frac{1}{x\left (\log (x)\right )^2}dx$ we use the substitution $u=\log (x)$.
   
   We have the following: \begin{align*}\int_2^b\frac{1}{x\left (\log (x)\right )^2}dx & =\int_{\log(2)}^{\log(b)}\frac{1}{e^u\left (\log (e^u)\right )^2}\cdot e^udu \\ & =\int_{\log(2)}^{\log(b)}u^{-2}du \\ & =-\left [u^{-1}\right ]_{\log (2)}^{\log (b)} \\ & =-\left (\frac{1}{\log (b)}-\frac{1}{\log (2)}\right ) \\ & =-\frac{1}{\log (b)}+\frac{1}{\log (2)}\end{align*}
   
   So, the integral is \begin{equation*}\lim_{b\rightarrow \infty}\int_2^b\frac{1}{x\left (\log (x)\right )^2}dx=\lim_{b\rightarrow \infty}\left (-\frac{1}{\log (b)}+\frac{1}{\log (2)}\right )=-0+\frac{1}{\log (2)}=\frac{1}{\log (2)}\in \mathbb{R}\end{equation*}
   
   Therefore, the integral $\displaystyle{\int_2^{\infty}\frac{1}{x\left (\log (x)\right )^2}dx}$ converges. Could we show the convergence also without calculating the integral? (Wondering)
   
   $$$$
   
2. $\displaystyle{\int_0^{\infty}e^{sx}\cos (tx)dx}$, wobei $s<0$ und $t\in \mathbb{R}$
   
   We have that \begin{equation*}\left |e^{sx}\cos (tx)\right |=\left |e^{sx}\right |\cdot \left |\cos (tx)\right |\leq \left |e^{sx}\right |=e^{sx}\end{equation*}
   
   We have that: \begin{equation*}\int_0^{\infty}e^{sx}=\lim_{b\rightarrow \infty}\int_0^b e^{sx}dx\end{equation*}
   
   Therfore we get \begin{equation*}\lim_{b\rightarrow \infty}\int_0^b e^{sx}dx=\lim_{b\rightarrow \infty}\left [\frac{1}{s}e^{sx}\right ]_0^b=\frac{1}{s}\lim_{b\rightarrow \infty}\left [e^{sx}\right ]_0^b=\frac{1}{s}\lim_{b\rightarrow \infty}\left (e^{sb}-1\right )=\frac{1}{s}\left (0-1\right )=-\frac{1}{s}\end{equation*}
   
   So, we have that \begin{equation*}\int_0^{\infty}e^{sx}=-\frac{1}{s}\in \mathbb{R}\end{equation*}
   
   That means that the integral $\displaystyle{\int_0^{\infty}e^{sx}dx}$ coverges and so the integral $\displaystyle{\int_0^{\infty}e^{sx}\cos (tx)dx}$ also converges.
   
   Is this correct? (Wondering)
   
   $$$$
   
3. $\displaystyle{\int_0^1\left (\log (x)\right )^2dx}$
   
   Calculating that integral we get $0$. So, the integral converges.
   
   Is there also an other way without calculating it? (Wondering)
   
   $$$$
   
4. $\displaystyle{\int_1^{\infty}\frac{x\sqrt{x}}{(2x+1)^2}dx}$
   
   Could you give a hint for that one? (Wondering)

 

[sarrah1]

[*] $\displaystyle{\int_1^{\infty}\frac{x\sqrt{x}}{(2x+1)^2}dx}$

I think it's divergent, since $x>\sqrt{x}$the integral by comparison is greater than

$\frac{1}{2}\int_{1}^{\infty} \,\frac{1}{2x+1}dx-\frac{1}{2}\int_{1}^{\infty} \,\frac{1}{{(2x+1)}^{2}}dx$

the first diverges

 

[mathmari]

[*] $\displaystyle{\int_1^{\infty}\frac{x\sqrt{x}}{(2x+1)^2}dx}$

I think it's divergent, since $x>\sqrt{x}$the integral by comparison is greater than

$\frac{1}{2}\int_{1}^{\infty} \,\frac{1}{2x+1}dx-\frac{1}{2}\int_{1}^{\infty} \,\frac{1}{{(2x+1)}^{2}}dx$

the first diverges

I see! (Nerd)

3. $\displaystyle{\int_0^1\left (\log (x)\right )^2dx}$

Calculating that integral we get $0$. So, the integral converges.

Is there also an other way without calculating it? (Wondering)

For that one, could we do the following:

We have that $\log (x)\leq x-1$ then $\log^2(x)\leq (x-1)^2$.

We have that $\int_0^1(x-1)^2dx=\ldots =\frac{1}{3}\in \mathbb{R}$. So, the integral of $(x-1)^2$ converges, and from the direct comparison we get that the integral of $\log^2(x)$ converges.

Is this correct? (Wondering)