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I don't understand why on the right-hand side we have $d\ln x$ and why the fraction disappeared. I have never seen this technique before. What is the logic behind this, I don't understand why the equivalence holds?

$$\int_{e}^{+\infty} \frac{(\ln x)^{-100}}{x}\, dx=\int_{e}^{+\infty}(\ln x)^{-100} \,d\ln x$$

DMcMor
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timtam
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  • You can do $u = \ln x$, then $\mathrm{d}u = \frac{1}{x} \mathrm{d}{x}$. Also take a look at this – talbi Dec 09 '20 at 21:17
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    $d (\ln x) = \frac{dx}{x}$. A style note: Do you see why the limits of the integral are irrelevant to the problem? Thus it is best to omit them, as you're more likely to get help. For the same reason, do you see why $(\ln x)^{-100}$ is irrelevant to your problem? Thus it is best to leave it out. When you do these obvious and immediate simplifications, you're more likely to see to the core of your problem (and perhaps solve it yourself). You're certainly more likely to garner assistance. – David G. Stork Dec 09 '20 at 21:17
  • @DavidG.Stork So in other words, it is best practice to state a question as general as possible, did I understand you correctly? – timtam Dec 09 '20 at 21:51
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    Sort of. "Pare a question to its basics, its essence, only the relevant matters." Go to this page and search "pare question to its basics." https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question – David G. Stork Dec 09 '20 at 21:55
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    @David G. Stork I agree with your idea in general, but in this case, if the OP doesn't know why the identity holds, it's understandable they might think that somehow the integral, the bounds, or the multiplying function might be relevant. Wouldn't you agree? – Diffusion Dec 10 '20 at 06:02
  • @Zachary: If the OP doesn't think about it, then sure... it is "understandable" (s)he might mistakenly think the integral limits might be relevant. But first things first: let's agree that they are not relevant here. I'm urging this OP (and those in the future) to think more deeply about a question before they pose it. Imagine putting the limits as some other values. Given that the are the same on the left & right sides of the equation, the OP realizes they are not relevant... or at least explore this. At the end the question becomes $\int \frac{dx}{x} = \int d(ln x)$. Helpful! – David G. Stork Dec 10 '20 at 16:48

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