Consider the following integral: I=\int_0^\infty \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}dx

kramtus51 2022-01-06 Answered
Consider the following integral:
\(\displaystyle{I}={\int_{{0}}^{\infty}}{\frac{{{x}-{1}}}{{\sqrt{{{2}^{{x}}-{1}}}{\ln{{\left({2}^{{x}}-{1}\right)}}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Ronnie Schechter
Answered 2022-01-07 Author has 54 answers
Sub \(\displaystyle{u}={\log{{\left({2}^{{x}}-{1}\right)}}}\). Then \(\displaystyle{x}=\frac{{\log{{\left({1}+{e}^{{u}}\right)}}}}{{\log{{2}}}},\ {\left.{d}{x}\right.}={\left(\frac{{1}}{{\log{{2}}}}\right)}{\left({d}\frac{{u}}{{{1}+{e}^{{-{u}}}}}\right.}\). The integral then becomes
\(\displaystyle{\frac{{{1}}}{{{\log{{2}}}}}}{\int_{{-\infty}}^{{\infty}}}{\frac{{{d}{u}}}{{{1}+{e}^{{-{u}}}}}}{\frac{{{\frac{{{\log{{\left({1}+{e}^{{u}}\right)}}}}}{{{\log{{2}}}}}}-{1}}}{{{u}}}}={\frac{{{1}}}{{{2}{{\log}^{{2}}{2}}}}}\)
\(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{d}{u}}}{{{\text{cosh}{{\left(\frac{{u}}{{2}}\right)}}}}}}{\frac{{{\log{{\left({1}+{e}^{{u}}\right)}}}-{\log{{2}}}}}{{{u}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}{{\log}^{{2}}{2}}}}}{\int_{{-\infty}}^{{0}}}{\frac{{{d}{u}}}{{{\text{cosh}{{\left(\frac{{u}}{{2}}\right)}}}}}}{\frac{{{\log{{\left({1}+{e}^{{u}}\right)}}}-{\log{{2}}}}}{{{u}}}}+{\frac{{{1}}}{{{2}{{\log}^{{2}}{2}}}}}{\int_{{0}}^{\infty}}{\frac{{{d}{u}}}{{{\text{cosh}{{\left(\frac{{u}}{{2}}\right)}}}}}}{\frac{{{\log{{\left({1}+{e}^{{u}}\right)}}}-{\log{{2}}}}}{{{u}}}}\)
The nasty pieces of the integral cancel, and we are left with
\(\displaystyle{\frac{{{1}}}{{{2}{{\log}^{{2}}{2}}}}}{\int_{{0}}^{\infty}}{\frac{{{d}{u}}}{{{\text{cosh}{{\left(\frac{{u}}{{2}}\right)}}}}}}={\frac{{\pi}}{{{2}{{\log}^{{2}}{2}}}}}\)
as correctly conjured.
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enhebrevz
Answered 2022-01-08 Author has 5005 answers
\(\displaystyle{I}={\int_{{0}}^{\infty}}{\frac{{{x}-{1}}}{{\sqrt{{{2}^{{x}}-{1}}}{\ln{{\left({2}^{{x}}-{1}\right)}}}}}}{\left.{d}{x}\right.}\)
With the change of variables \(\displaystyle{z}\equiv{2}^{{x}}-{1}\to{x}=\frac{{\ln{{\left({1}+{z}\right)}}}}{{\ln{{\left({2}\right)}}}}\), I is reduced to
\(\displaystyle{I}={\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{\infty}}{\frac{{{\ln{{\left({1}+{z}\right)}}}-{\ln{{\left({2}\right)}}}}}{{{z}^{{\frac{{1}}{{2}}}}{\left({1}+{z}\right)}{\ln{{\left({z}\right)}}}}}}{\left.{d}{z}\right.}\)
Now, we split the integral from (0,1) and from \(\displaystyle{\left({1},\infty\right)}\). In the second one, we makes the change \(\displaystyle{z}\to\frac{{1}}{{z}}\) such that we are left with an integration over \(\displaystyle{\left({0},{1}\right)}\):
\(\displaystyle{I}={\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{z}\right)}}}-{\ln{{\left({2}\right)}}}}}{{{z}^{{\frac{{1}}{{2}}}}{\left({1}+{z}\right)}{\ln{{\left({z}\right)}}}}}}{\left.{d}{z}\right.}+{\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+\frac{{1}}{{z}}\right)}}}-{\ln{{\left({2}\right)}}}}}{{{z}^{{-\frac{{1}}{{2}}}}{\left({1}+\frac{{1}}{{z}}\right)}{\left[-{\ln{{\left({z}\right)}}}\right]}}}}{\frac{{{\left.{d}{z}\right.}}}{{{z}^{{2}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{z}\right)}}}-{\ln{{\left({2}\right)}}}}}{{{z}^{{\frac{{1}}{{2}}}}{\left({1}+{z}\right)}{\ln{{\left({z}\right)}}}}}}{\left.{d}{z}\right.}-{\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}\)
\(\displaystyle{\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{z}\right)}}}-{\ln{{\left({z}\right)}}}-{\ln{{\left({2}\right)}}}}}{{{z}^{{\frac{{1}}{{2}}}}{\left({1}+{z}\right)}{\ln{{\left({z}\right)}}}}}}{\left.{d}{z}\right.}\)
\(\displaystyle={\frac{{{1}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{{1}}}{\frac{{{1}}}{{{z}^{{\frac{{1}}{{2}}}}{\left({1}+{z}\right)}}}}{\left.{d}{z}\right.}\),
\(\displaystyle={\frac{{{2}}}{{{{\ln}^{{2}}{\left({2}\right)}}}}}{\int_{{0}}^{{1}}}{\frac{{{d}{r}}}{{{r}^{{2}}+{1}}}}={\frac{{\pi}}{{{2}{{\ln}^{{2}}{\left({2}\right)}}}}}\)
\(\displaystyle{\arctan{{\left({1}\right)}}}={\frac{{\pi}}{{{4}}}}\)
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karton
Answered 2022-01-11 Author has 8659 answers

Substitute \((2^x-1)=t^2\) to get,
\(I=\frac{1}{\ln^2 2}\int_0^\infty(\frac{\ln(t^2+1)-\ln2}{(t^2+1)\ln t})dt\)
Substitute \(t\mapsto\frac{1}{t}\)
\(\to I=-\frac{1}{\ln^2 2}\int_0^{\infty}(\frac{\ln(t^2+1)-\ln2}{(t^2+1)\ln t}dt+\frac{2}{\ln^2 2}\int_0^\infty\frac{dt}{t^2+1} \\\to I=-I+\frac{\pi}{\ln^2 2} \\\to I=\frac{\pi}{2\ln^2 2}\)

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