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I have to show that the irregular integral given by

$$ \int_a^\infty \frac{\ln(x)}{x^s} dx $$ is convergent for $a > 0 $ and to calculate its value where $s$ is a real number $>1$.

In an earlier question I have calculated the above integral as

$$ \int \frac{\ln(x)}{x^s}dx = \ln(x) \cdot \frac{x^{-s+1}}{-s+1} - \frac{x^{-s+1}}{(-s+1)^2} $$ but I am not sure if this is of any help.

Can you guide me in the right direction of where to start?

Thanks in advance.

Bernard
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Mathias
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  • You have the antiderivative, so by the fundamental theorem of calculus,$$\int_a^\infty\frac{\ln x}{x^s},\mathrm dx=\lim_{t\to\infty}\left(\frac{t^{1-s}\ln t}{1-s}-\frac{t^{1-s}}{(1-s)^2}\right)-\frac{a^{1-s}\ln a}{1-s}+\frac{a^{1-s}}{(1-s)^2}$$ – user170231 Mar 17 '20 at 14:41
  • So this means that the integral is convergent for $a > 0$? We could not have done this if $a<0$? – Mathias Mar 17 '20 at 14:48
  • Also relevant: https://math.stackexchange.com/a/364169 – an4s Mar 17 '20 at 14:48
  • Is it because ln(x) is not defined for x<0? – Mathias Mar 17 '20 at 14:49

2 Answers2

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Hint:

Use asymptotic analysis:

$$\frac{\ln x}{x^s}=o\Bigl(\frac 1{x^t}\Bigr)\quad\text{ for any }t<s $$

and the (improper) integral of the latter function is convergent if $t>1$.

Bernard
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First, since we are trying to determine whether the limit of $$\int_a^b\frac {\ln x}{x^s}dx$$ exists as $b \to \infty,$ we may as well start the integral at 1, since the difference between $$\int_a^b\frac {\ln x}{x^s}dx \text { and } \int_1^b\frac {\ln x}{x^s}dx $$ is just the fixed number $$\int_a^1\frac {\ln x}{x^s}dx \text { or } \int_1^a\frac {\ln x}{x^s}dx $$ depending on whether $a<1$ or $a>1.$ Then we have $$\int_1^b \frac{\ln(x)}{x^s}dx = \ln(x) \cdot \frac{x^{-s+1}}{-s+1}|_1^b +\int_1^b \frac{1}{(s-1)x^s}dx$$ $$= -\ln(b) \cdot \frac{b^{-s+1}}{s-1} +\int_1^b \frac{1}{(s-1)x^s}dx$$ $$\le \int_1^b \frac{1}{(s-1)x^s}dx$$ Since $\frac{\ln x}{x^s} \ge 0 $ on $[1,\infty)$ and since $$\int_1^{\infty} \frac{1}{(s-1)x^s}dx$$ converges, so does the given integral.

P. Lawrence
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  • Can I show that limit exists as I earlier found that $\int \frac{\ln(x)}{x^s}dx = \ln(x) \cdot \frac{x^{-s+1}}{-s+1} - \frac{x^{-s+1}}{(-s+1)^2}$ so by the fundamental theorem of calculus the limit $\lim_{b \rightarrow \infty} \int_a^b \frac{\ln(x)}{x^s}$ does exist? – Mathias Mar 17 '20 at 21:22