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badreferences

  • 4 years ago

\[\int\frac{1}{\ln x}\,dx\]

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  1. badreferences
    • 4 years ago
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    Now, I'm pretty sure\[u=\ln x\]\[1=\frac{dx}{du}\frac{1}{x}\]\[dx=x\,du=e^{\ln x}\,du=e^u\,du\]\[\int\frac{e^u}{u}\,du\]And I could swear that this can be expressed as some disgusting power series variant.

  2. badreferences
    • 4 years ago
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    \[\int\left(\frac{1}{u}e^u\right)\,du=\int\left(\frac{1}{u}\sum_{n=1}^\infty u^n\right)\,du\]

  3. badreferences
    • 4 years ago
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    @Zarkon Check this out please. :3

  4. badreferences
    • 4 years ago
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    \[\int\sum_{n=1}^\infty u^{n-1}\,du\]

  5. badreferences
    • 4 years ago
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    Whoops, forgot something!

  6. Zarkon
    • 4 years ago
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    there is no antiderivative in terms of elementary functions

  7. badreferences
    • 4 years ago
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    Expressing the answer in terms of a sum would be fine.

  8. badreferences
    • 4 years ago
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    \[\int\frac{1}{u}e^u\,du=\int\frac{1}{u}\sum_{n=1}^\infty\frac{u^n}{n!}\,du=\int\sum_{n=1}^\infty\frac{u^{n-1}}{n!}\,du=\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}\]

  9. badreferences
    • 4 years ago
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    @FoolForMath :D

  10. badreferences
    • 4 years ago
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    \[\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}=\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]

  11. badreferences
    • 4 years ago
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    Can it be further simplified?

  12. badreferences
    • 4 years ago
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    A better question: what does\[\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]converge to?

  13. badreferences
    • 4 years ago
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    By the ratio test it converges.

  14. Zarkon
    • 4 years ago
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    you should include your constant of integration along with the radious of convergence.

  15. mukushla
    • 3 years ago
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    just a correction :\[\int \frac{1}{u} \sum_{n=0}^{\infty} \frac{u^n}{n!} du=\int (\frac{1}{u}+\sum_{n=1}^{\infty} \frac{u^{n-1}}{n!})\ du=\ln u+\sum_{n=1}^{\infty} \frac{u^{n}}{n.n!}+C\]

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