I am currently stuck with this integral and I have never seen such combination before. I've tried to divide it into separate integrals and it starts to diverge then. I've tried to complete the square and play around it, but it does not work either. Can you please give me some hints on how to approach such integral? Maybe I just don't see some obvious move?
$$\int_{0}^{1}\frac{e^{2y}(4y^2-4y+2)-2}{y^2}dy$$
On
Observe
$$\int_{0}^{1}\frac{e^{2y}(4y^2-4y+2)-2}{y^2}dy=\int_{0}^{1}e^{2y}\frac{4y^2-4y+2-2e^{-2y}}{y^2}dy$$
and
$$2e^{-2y}=2\sum_{n=0}^{\infty}\frac{(-2y)^n}{n!}=2-4y+4y^2+2\sum_{n=3}^{\infty}\frac{(-2y)^n}{n!}$$ and then $$\int_{0}^{1}\frac{e^{2y}(4y^2-4y+2)-2}{y^2}dy=\int_{0}^{1}-e^{2y}2\sum_{n=3}^{\infty}\frac{(-2)^n}{n!}y^{n-3}dy=\sum_{n=0}^{\infty}\frac{(-2)^{n+4}}{(n+3)!}\int_0^1 e^{-2y}y^ndy$$
Later integral is a incomplete gamma and then
$$\int_{0}^{1}\frac{e^{2y}(4y^2-4y+2)-2}{y^2}dy=\sum_{n=0}^{\infty}\frac{(-1)^{n}8}{(n+3)!}\gamma(n+1,2)$$
I finally figured out how to do it. Though I really appreciate popi's answer :)
Here how it goes:
$$\int_{0}^{1}\frac{e^{2y}(4y^2-4y+2)-2}{y^2}dy =\quad 4\int_{0}^{1}{e^{2y}dy + \int_{0}^{1}{\frac{-4e^{2y}+2e^{2y}-2}{y^2}dy} = 2(e^2-1) + 2\int_{0}^{1}\frac{e^{2y}(1-2y)-1}{y^2}dy=...}$$
Then I made a little substitution:
$${t=e^{2y}\quad dy=\frac{dt}{2t}\quad y=\frac{lnt}{2}}$$
Hence:
$${...=2(e^2-1)+4\int_{1}^{e^2}\frac{t(1-lnt)-1}{tln^2t}dt=...}$$
After some rearrangement:
$${...=2(e^2-1)+4\Bigg(\int_{1}^{e^2}\frac{t-1}{tln^2t}dt-\int_{1}^{e^2}\frac{1}{lnt}dt\Bigg)=...}$$
Then I integrated by parts with ${u=1-t}$ and ${dv=-\frac{dt}{tln ^2t}}$ and got:
$${...=2(e^2-1)+4\Bigg(\Bigg(\frac{1-t}{lnt}\Bigg)\Bigg|^{e^2}_1+\int_{1}^{e^2}\frac{dt}{lnt}-\int_{1}^{e^2}\frac{dt}{lnt}\Bigg)}=$$
$${2(e^2-1)+4\Big(\frac{1-e^2}{2}+1\Big)=4}$$