Integral of \int\frac{1}{\log x}dx

Lainey Goodwin

Lainey Goodwin

Answered question

2022-01-22

Integral of 1logxdx

Answer & Explanation

fumanchuecf

fumanchuecf

Beginner2022-01-23Added 21 answers

If you don't like special functions symbols:
1logxdx=ln10lnxdx
Let u=lnx,
Then x=eu
dx=eudu
ln10lnxdx
=euln10udu
=n=0unln10n!udu
=n=0un1ln10n!du
=(ln10u+n=1un1ln10n!)du
=ln10lnu+n=1unln10n!n+C
=ln10lnlnx+n=1(lnx)nln10n!n+C

Georgia Ingram

Georgia Ingram

Beginner2022-01-24Added 11 answers

I assume that \(\displaystyle{\log{{x}}}\) denotes the natural logarithm.
The antiderivative you are analyzing has no expression in terms of “elementary function”, just like the case of the antiderivative of \(\displaystyle{e}^{{−{x}^{{2}}}}\).
You can transform it in various ways; with the substitution \(\displaystyle{t}=−{\log{{x}}}\), we have \(\displaystyle{x}={e}−{t}\) and \(dx=−e^{−t}dt\), so we obtain
\(\displaystyle\int{\frac{{{e}^{{−{t}}}}}{{{t}}}}{\left.{d}{t}\right.}\)
but this doesn't help much.

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