Problem with divergent integral

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SUMMARY

The discussion centers on the evaluation of the improper integral \(\int_{a}^{\infty} \left( \dfrac{1}{t} - \dfrac{1}{t-1} \right) ~ dt\) for \(a > 1\). The correct evaluation yields \(- \log \left| \dfrac{a}{a-1} \right|\). The confusion arises when attempting to decompose the integral into two parts, leading to an indeterminate form of \(\infty - \infty\). The mistake lies in assuming that both parts of the decomposed integral are finite, which is not the case.

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parton
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I'm confused with the following integral.

Let a > 1.

[tex]\int_{a}^{\infty} \left( \dfrac{1}{t} - \dfrac{1}{t-1} \right) ~ dt = \left[ \log \left| \dfrac{t}{t-1} \right| \right]_{a}^{\infty} = - \log \left| \dfrac{a}{a-1} \right|[/tex]

This should be the correct result. But I could also decompose the integral into two parts (because the integrand is a sum) and compute:

[tex]\int_{a}^{\infty} \left( \dfrac{1}{t} - \dfrac{1}{t-1} \right) ~ dt = \left| \log \vert t \vert \right|_{a}^{\infty} - \left[ \log \vert t - 1 \vert \right]_{a}^{\infty} = - \log \left| \dfrac{a}{a-1} \right| + \infty - \infty[/tex]

But [tex]\infty - \infty[/tex] is of course not defined!

Where did I make a mistake? I don't find it.
 
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Decomposing the integral into two parts is justified only if both integrals are finite.
 
Thanks :smile:
 

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