Here's the question you clicked on:
badreferences
\[\int\frac{1}{\ln x}\,dx\]
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.
\[\int\left(\frac{1}{u}e^u\right)\,du=\int\left(\frac{1}{u}\sum_{n=1}^\infty u^n\right)\,du\]
@Zarkon Check this out please. :3
\[\int\sum_{n=1}^\infty u^{n-1}\,du\]
Whoops, forgot something!
there is no antiderivative in terms of elementary functions
Expressing the answer in terms of a sum would be fine.
\[\int\frac{1}{u}e^u\,du=\int\frac{1}{u}\sum_{n=1}^\infty\frac{u^n}{n!}\,du=\int\sum_{n=1}^\infty\frac{u^{n-1}}{n!}\,du=\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}\]
\[\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}=\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]
Can it be further simplified?
A better question: what does\[\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}\]converge to?
By the ratio test it converges.
you should include your constant of integration along with the radious of convergence.
just a correction :\[\int \frac{1}{u} \sum_{n=0}^{\infty} \frac{u^n}{n!} du=\int (\frac{1}{u}+\sum_{n=1}^{\infty} \frac{u^{n-1}}{n!})\ du=\ln u+\sum_{n=1}^{\infty} \frac{u^{n}}{n.n!}+C\]