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I have to find out for which $b$, the improper integral exist.

$$\int_e^\infty \! \frac{(\ln (x))^b}{x}\ \, \mathrm{d}x$$

I know that the improper integral exists if the limit exist.

Is there a general methodic to find these $b$ or just trying?

Thx in advance

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    Start by solving the integral with $u=\ln(x)$. – 2017-01-28

3 Answers 3

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Hint For $b=-1$ a primitive of the function is $\ln(\ln x)$ and so we see the integral is divergent. By comparison we get also the divergence for $b>-1$. Now let $b<-1$, a primitive is $\frac1{b+1}(\ln x)^{b+1}$ and in this case the integral is convergent.

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This is called Bertrand's Integral.

It's general form is given by, $$\int_e^{+\infty} \frac{1}{x^\alpha\log(x)^\beta} dx $$ and it converges if $(\alpha>1)$ or $(\alpha=1,\beta>1)$.

Hence, $\displaystyle\int_e^{+\infty}\frac{\log(x)^b}{x}dx $ converges for $b<-1$.

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One can first solve the integral with the substitution $u=\ln(x)$, which gives us

$$\int u^b\ du=\frac{u^{b+1}}{b+1}\color{#888888}{+c}$$

Placing in bounds, we then get

$$\left.\int_e^a\frac{(\ln(x))^b}x\ dx=\frac{(\ln(x))^{b+1}}{b+1}\right|_{x=e}^{x=a}$$

as $a\to\infty$, this integral exists if $(\ln(x))^{b+1}$ goes to $0$, i.e. for $b<-1$.

For $b=-1$, the integral turns out to be $\ln(\ln(x))$, which diverges.

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    Thanks! But I don't understand the step with Placing in bounds. Whre's the x in the bottom of the fraction? We just substituted u=ln(x), or not? Would be fine if you explain the substitution – 2017-01-28
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    @Jones876 It is just a standard substitution: $u=\ln(x)$ and $du=\frac1x\ dx$. That's where the $\frac1x$ went. – 2017-01-28
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    Oh okay now I understand. Does it converges for b<-1 because of the Harmonic progression? – 2017-01-28
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    @Jones876 It converges there because $\lim_{a\to\infty}$ exists. – 2017-01-28