## badreferences 3 years ago $\int\frac{1}{\ln x}\,dx$

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!

6. Zarkon

there is no antiderivative in terms of elementary functions

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

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

@FoolForMath :D

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

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.

14. Zarkon

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

15. mukushla

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