# What is the integral of x/ln(x)?

I need calculate this integral :

$$\int_{e}^{2e} \frac{x}{\ln(x)} dx$$

But I don't know how, can you help me please? Thank you !

• English isn't an issue, but you should read the guide to asking good questions. – user147263 Jul 18 '14 at 15:27
• There is no elementary antiderivative. – André Nicolas Jul 18 '14 at 15:28
• – Aryabhata Jul 18 '14 at 15:29

## 4 Answers

You have: $$\int_{e}^{2e}\frac{x}{\log x}dx = e^2\int_{1}^{2}\frac{x}{1+\log x}dx=e^2\int_{0}^{\log 2}\frac{e^{2x}}{1+x}dx=\int_{1}^{1+\log 2}\frac{e^{2x}}{x}dx.$$ Despite the fact that no elementary antiderivative exists, you can exploit the fact that $\frac{e^{2x}}{1+x}$ is a very regular function on $[0,\log 2]$, and integrate termwise its Taylor series in $x=0$. We have: $$\begin{eqnarray*}\int_{1}^{1+\log2}\frac{e^{2x}}{x}dx &=& \log(1+\log 2)+\int_{1}^{1+\log 2}\frac{e^{2x}-1}{x}dx\\&=&\log(1+\log 2)+\left[\sum_{j=1}^{+\infty}\frac{(2x)^{j}}{j\cdot j!}\right]_{1}^{1+\log 2}\\&=&\log(1+\log 2)+\sum_{j=1}^{+\infty}\frac{2^j((1+\log 2)^j-1)}{j\cdot j!}\\&=&\log(1+\log 2)+\sum_{j=1}^{+\infty}\frac{2^j}{j\cdot j!}\sum_{k=1}^{j}(\log 2)^k\\&=&\log(1+\log 2)+\sum_{k=1}^{+\infty}(\log 2)^k\sum_{j\geq k}\frac{2^j}{j\cdot j!}.\end{eqnarray*}$$ Another efficient technique to calculate such an integral numerically is to exploit the continued fraction representation for the exponential integral.

This is not an elementary integral. Let $x=e^{-z}$ and observe that the integral be written in terms of the exponential integral, defined as $$\text{Ei}(t) = -\int_{-t}^\infty \frac{e^{-z}}{z}\,dz$$ which as noted on the Wikipedia page is not an elementary function.

The indefinite integral cannot be expressed in terms of elementary functions. The integral is, quite unsatisfactorily, expressed in terms of the exponential integral $\mathrm{Ei}(x)$. We have $$\int \frac{x}{\ln x}~\mathrm{d}x = \mathrm{Ei}(2\ln x)+C$$

The change of variables $x=e^y$ gives the integral $\int \frac{e^{2y}}ydy$, which is known not to be computable in elementary functions. Thus, the answer can be only expressed via the exponential integral, $\int...=Ei(2\ln2\epsilon)-Ei(2\ln\epsilon)$.