Question

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

Answer 1

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.

Answer 2

\[\int\left(\frac{1}{u}e^u\right)\,du=\int\left(\frac{1}{u}\sum_{n=1}^\infty u^n\right)\,du\]

Answer 3

@Zarkon Check this out please. :3

Answer 4

\[\int\sum_{n=1}^\infty u^{n-1}\,du\]

Answer 5

Whoops, forgot something!

Answer 6

there is no antiderivative in terms of elementary functions

Answer 7

Expressing the answer in terms of a sum would be fine.

Answer 8

\[\int\frac{1}{u}e^u\,du=\int\frac{1}{u}\sum_{n=1}^\infty\frac{u^n}{n!}\,du=\int\sum_{n=1}^\infty\frac{u^{n-1}}{n!}\,du=\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}\]

Answer 9

@FoolForMath :D

Answer 10

\[\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}=\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]

Answer 11

Can it be further simplified?

Answer 12

A better question: what does\[\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]converge to?

Answer 13

By the ratio test it converges.

Answer 14

you should include your constant of integration along with the radious of convergence.

Answer 15

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