How does Wolfram Alpha get $\displaystyle\int\dfrac y{\ln y}dy\gets\operatorname{Ei}(-2\ln y)+c$?

Question

Now this goes back when I was trying to solve the differential equation$$y'y=\ln y$$which is a first order nonlinear ODE. Now, I was able to get this integral out of it:$$\int\dfrac y{\ln y}dy=x+c_0$$however it was there that I got stuck, since even though I know that I could probably evaluate the integral using a change of variables, I'm not sure which one to use. I had originally tried $u=\ln y$, which then $e^u=y\implies e^udu=dy$ and then the integral becomes$$\int e^u\cdot\dfrac{e^u}{u}du=\int\dfrac{e^{2u}}{u}du$$although now I don't know how to continue from here. WA gives$$\int\dfrac y{\ln y}=\operatorname{Ei}(-2\ln y)+c$$but I'm unsure how it gets that result. So my question is:

How does WA get $\int y/\ln(y)dy=\operatorname{Ei}(-2\ln y)+c$?

Answer 1

Letting $u=2\ln y$ yields $$ \begin{aligned} I & =\int \frac{e^{\frac{u}{2}}}{\frac{u}{2}} \cdot \frac{1}{2} e^{\frac{u}{2}} d u \\ & =\int \frac{e^u}{u} d u \\ & =E_i(u) +C\\ & =E_i(2 \ln y)+C \end{aligned} $$ where $E_i(.)$ is the Exponential Integral. The answer is different from OP’s and checked by WA .

Answer 2

It turns out from here, we can substitute $2u=r$ to turn the integral into$$\int\require{cancel}\cancel{\dfrac12\cdot2}\dfrac{e^r}rdr\gets\operatorname{Ei}(-r)+c_1$$and then undoing the substitutions gives us$$\operatorname{Ei}(-r)+c_1\gets\operatorname{Ei}(-2u)+c_1\gets\operatorname{Ei}(-2\ln y)+c_1$$

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Notes on the diff. eq. (will remove if not needed)

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As for the differential equation, it is not possible to isolate $y(x)$ due to $\operatorname{Ei}(x)$ not having an inverse that is representable using standard mathematical functions, however if we were to define $h(x)$ being the supposed inverse of $\operatorname{Ei}(x)$, the solution would be$$y(x)=e^{-0.5h(x+c_0+c_1)}$$however, it would probably just be best to say that the solution to the diff. eq. is just $x+c_0=\operatorname{Ei}(-2\ln y)$ again due to $\operatorname{Ei}(x)$ having no inverse in terms of standard mathematical functions.

Answer 3

$$y'y=\log(y)$$ Switch variables

$$\frac y{x'}=\log(y) \quad \implies \quad x'=\frac{y}{\log (y)}$$ $$x+c=\text{Ei}(2 \log (y))$$

Answer 4

Let u=Iny. du=dy/y. dy=ydu. y=e^u $ (e^u)/u × (e^u du) $ (e^2u)/u × du. Let v=2u. dv=2du. U=v/2 $ (e^v)/(v/2) × dv/2 $ (e^v)/v × dv = Ei(v)+C. But v=2u and u=Iny $ y/Iny × dy =Ei(2Iny)+C