Help solving improper integral

Question

I would like to know how could I solve (in terms of $x$ and $a$)

$$\int_{x}^{\infty} \frac{a(1+\ln{y})}{y\ln^2{y}} dy$$

for $x>0$ ; $a= constant$ ; $-1<a<1$ ?

Thanks.

Answer 1

Hint

Start changing variable $y=e^t$ to get $$I=\int\frac{1+\ln{y}}{y\ln^2{y}} dy=\int \frac{t+1}{t^2}\,dt=\log (t)-\frac{1}{t}$$ and see what happens if $t \to \infty$.

Answer 2

I will assume for simplicity that $x>1$, so that $\ln(x)>0$. Since $$ \frac{1+\ln(y)}{y\ln(y)^2}>\frac{\ln(y)}{y\ln(y)^2}=\frac{1}{y\ln(y)} $$ for all $y\geq x$, and since $$ \int_x^{\infty}\frac{dy}{y\ln(y)}=\int_{\ln(x)}^{\infty}\frac{du}{u}=\infty$$ it follows that your integral diverges, at least when $a\neq 0$.

Answer 3

For the indefinite integral, substitute $\text{u}=\ln\left(\text{y}\right)$ and $\text{d}\text{u}=\frac{1}{\text{y}}\space\text{d}\text{y}$:

$$\mathcal{I}\left(\text{a},\text{y}\right)=\int\frac{\text{a}\left(1+\ln\left(\text{y}\right)\right)}{\text{y}\ln^2\left(\text{y}\right)}\space\text{d}\text{y}=\text{a}\int\frac{1+\text{u}}{\text{u}^2}\space\text{d}\text{u}=\text{a}\left\{\int\frac{1}{\text{u}^2}\space\text{d}\text{u}+\int\frac{1}{\text{u}}\space\text{d}\text{u}\right\}=$$ $$\text{a}\left(-\frac{1}{\text{u}}+\ln\left|\text{u}\right|\right)+\text{K}=\text{a}\left(\ln\left|\ln\left(\text{y}\right)\right|-\frac{1}{\ln\left(\text{y}\right)}\right)+\text{K}$$

Now, for the lower bound:

$$\mathcal{I}\left(\text{a},\text{x}\right)=\text{a}\left(\ln\left|\ln\left(\text{x}\right)\right|-\frac{1}{\ln\left(\text{x}\right)}\right)$$

And, for the upper bound:

$$\lim_{\text{y}\to\infty}\mathcal{I}\left(\text{a},\text{y}\right)=\lim_{\text{y}\to\infty}\text{a}\left(\ln\left|\ln\left(\text{y}\right)\right|-\frac{1}{\ln\left(\text{y}\right)}\right)=\text{a}\left\{\lim_{\text{y}\to\infty}\left(\ln\left|\ln\left(\text{y}\right)\right|-\frac{1}{\ln\left(\text{y}\right)}\right)\right\}\space\to\infty$$