Mistake in evaluating $\int\frac{dx}{\ln(x)}$

Question

Evaluate: $$I=\int\dfrac{dx}{\ln(x)}$$

My attempt: $$$$ $$I=\int \dfrac{x'}{\ln(x)} dx$$Integrating by Parts,$$\dfrac{x}{\ln(x)}-\int\dfrac{x}{(\ln(x))'}dx$$$$=\dfrac{x}{\ln(x)}-\int x^2dx=\dfrac{x}{\ln(x)}-\dfrac{x^3}{3}$$$$$$

This answer of mine was deemed incorrect. However despite checking my attempt several times, I cannot understand where I have made a mistake. I would be truly grateful if somebody would kindly point out my error. Many thanks in advance!

PS. Kindly ignore C.Also, $\alpha'$ denotes the derivative of $\alpha$ with respect to $x$.

Answer 1

Your mistake is to assume that $$ \left(\frac1{\ln x}\right)'=\frac1{\left(\ln x\right)'} $$ which is $\color{red}{\text{wrong}}$.

Answer 2

When you integrate by parts, you should differentiate $1/\ln x$, and by the chain rule this yields

$$ - \frac{1}{x (\ln x)^2} $$

The integral you're trying to compute is called the logarithmic integral function, and you won't be able to find a 'simple' closed form for it. This integral has to do with the Prime Number Theorem.

Answer 3

$$I=\int\dfrac{dx}{\ln(x)}$$

$$u=\ln x,1=\frac{dx}{du}\frac{1}{x},dx=x\,du,dx=e^{\ln x}du,dx=e^u\,du$$

$$\int\frac{e^{u}}{u}\,du=\int\frac{1}{u}e^u\,du=\int\frac{1}{u}\sum_{n=0}^{\infty}\frac{u^n}{n!}\,du=\ln(u)+\int\sum_{n=1}^{\infty}\frac{u^{n-1}}{n!}\,du$$

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

$\Longrightarrow$

$$\int\frac{e^{u}}{u}\,du=\ln(u)+\sum^{\infty}_{n=1}\frac{u^{n}}{n!n}=\ln\bigl(\ln(x)\bigr)+\sum^{\infty}_{n=1}\frac{\ln(x)^{n}}{n!n}$$

Answer 4

You have $I = \int \frac{ d x}{\ln (x)} = \int \frac{ 1 }{ \ln (x) } dx$, where you write $I = \int \frac{ x'}{\ln (x)} dx$.

Furthermore, by the integration by parts you have $\int f g' dx = fg - \int f' gdx$ with $g' = x'$ and $f = \frac{1}{\ln(x)}$, so you do not have $f'= x$, but $f'= [\ln(x) ]' \frac{ 1 }{\ln^2(x)} = \frac{1}{x \ln^2(x)}$.

Finally, Wolfram Alpha says that $\int \frac{ 1}{\ln(x)} dx$ is a special function, so I am not sure if there is a nice formula as an answer.

Answer 5

You can try evaluating it with taylor series. Approximation at $e$: $\sum \frac{\frac{d^n}{d^nx}\int \frac{dx}{ln(e)}}{n!} (x-e)^n $ https://en.wikipedia.org/wiki/Logarithmic_integral_function