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I would like to ask some assistance with how to integrate : $\dfrac{x}{\ln x}$

I could really use an explanation and final answer.

Thank you very much.

Michael Albanese
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user118972
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  • You added the tag (definite-integral). In that case, which are the limits of integration? – Ian Mateus Jan 07 '14 at 13:44
  • Compute the power series expansion for $x/ln(x)$ pending the limits of integration, and then it's a simple integration. You can also change variables: $x=e^t$ to get an integral of the form: $e^{2t} / t$ which is more easy to expand in a power series. – MathematicalPhysicist Jan 07 '14 at 13:50
  • Thank you-- im sorry. between e^5 to e^6 – user118972 Jan 07 '14 at 13:51
  • About e2t/t , thank you for your answer but I don't seem to realize how to finish it from there, I ended up in a non-ending loop with that one – user118972 Jan 07 '14 at 13:53

2 Answers2

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Change the variable $x$ into $y=\ln x$, you get $$\int^x\frac{u}{\ln u}\mathrm{d}u=\int^{\ln x}\frac{\mathrm{e}^{2y}}y\mathrm{d}y.$$ This is an so-called exponential integral $\mathrm{Ei}$. Thus the primitive of $x\mapsto x/\ln x$ is $x\mapsto\mathrm{Ei}(2\ln x)$.

Tom-Tom
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Set $\,y:=x^2\,$ then for $x>0$ : \begin{align} \int \frac{x\,dx}{\ln(x)}&=\int \frac{2\,x\,dx}{2\ln(x)}\\ &=\int \frac{d(x^2)}{\ln(x^2)}\\ &=\int \frac{dy}{\ln(y)}\\ &=\operatorname{li}(y)\\ &=\operatorname{li}\left(x^2\right) \end{align} with $\operatorname{li}$ the logarithmic integral function (corresponding to V. Rossetto's answer).

Raymond Manzoni
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    $\mathrm{Ei}(x)=\mathrm{li}(\mathrm{e}^x)$ indeed ! It's good to remember this identity, since $\mathrm{Ei}$ is implemented in the GNU Scientific Library (GSL) but $\mathrm{li}$ is not ! – Tom-Tom Jan 07 '14 at 14:57
  • @V.Rossetto: I think too that $\mathrm{Ei}$ is more adapted for numerical evaluation than $\mathrm{li}$. One should merely take care that some implementations of the function '$\mathrm{Ei}$' implement in reality the related $\mathrm{E_1}$ function. – Raymond Manzoni Jan 07 '14 at 23:24