$$\int\dfrac{1}{e^\frac{2}{y}}dy$$
I'm trying to integrate $\frac{dy}{dt}=e^{2/y}$. I've separated the equation to $\frac{1}{e^{2/y}}dy=dt$ so that I can integrate each side with respect to the corresponding variables. Computing the integral of $dt$ is trivial but I'm confused on how to compute the integral of $\frac{1}{e^{2/y}}dy$.
On
Unfortunately, this function has no elementary antiderivative:
$$\int e^{-2/y}\,dy=2\,\Gamma\left( -1,\frac2y\right)+C$$
where $\Gamma(s,x)$ is the upper incomplete gamma function.
I arrived at this just by using integral-calculator.com.
(After @MyGlasses’s edit, I think his/hers is preferable. Nice trick, by the way!)
On
Considering $$I=\int e^{-2/y}\,dy$$ integrate by parts $$u'=dy\implies u=y$$ $$v=e^{-2/y}\implies v'=\frac{2 e^{-2/y}}{y^2}\,dy$$ making $$I=y\,e^{-2/y}-2\int \frac{ e^{-2/y}}{y}\,dy=y \,e^{-2/y} +2\, \text{Ei}\left(-\frac{2}{y}\right)$$ where appears the exponential integral function.
This is just another represnation of what Chase Ryan Taylor gave in his/her answer.
Hint:) This makes the integral simpler, with substitution $y=-\dfrac2u$ we have $dy=\dfrac{2}{u^2}$ and $$\int\dfrac{1}{e^\frac{2}{y}}dy=2\int\dfrac{e^u}{u^2}du$$
Edit: \begin{align} \int\dfrac{1}{e^\frac{2}{y}}dy &= \int1-\dfrac{2}{y}+\dfrac{2^2}{2!}\dfrac{1}{y^2}-\dfrac{2^3}{3!}\dfrac{1}{y^3}+\cdots dy \\ &= y-2\ln y-\dfrac{2^2}{2!}\dfrac{1}{y}+\dfrac{2^2}{3!}\dfrac{1}{y^2}+\cdots \end{align}