How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$? I tried integration by parts but failed. Then i tried to use infinite series.
Use $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
we get \begin{align}
I & =\int_{x^2}^{x} \frac{1}{y}+1+\frac{y}{2!}+\frac{y^2}{3!}+\cdots dy \\
 &=ln(\frac{1}{x})+(x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots
\end{align}
Now get back to $\int_0^1 I dx$. We need to find $P=\int_0^1 ln(\frac{1}{x})$ and $Q=\int_0^1 (x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots$
integrate by parts, we find $P=0$.
But for Q,we have $\left((\frac{1}{2}x^2-\frac{1}{3}x^3)+\frac{1}{4}(\frac{1}{3}x^3-\frac{1}{5}x^5)+ \cdots \right)_0^1$
Can we simplify the infinite sum in the bracket as a simple result?
 A: There is a theorem that says when you are given double integrals, you must flip them. Well, not always, but in this case you want to do that.
Reparametrize your region $R$ as $R = \{(x,y) | \sqrt{y} \le x \le y, 0 \le x \le 1\}.$
Then by Fubini, we can write 
$$\int_0^1 \int_{x^2}^x \frac{x}{y}e^y dydx = \int_0^1 \int_{\sqrt{y}}^y \frac{x}{y}e^y dxdy.$$
This is much easier to integrate. You end up having to integrate $e^y$ and $ye^y$. The former is immediate, the latter an integration by parts.
A: This is the Exponential integral, and it can't be expressed in terms of standard functions. However, one can first integrate over $x$ (with slightly more complicated bounds), and then the integration in $y$ becomes much more straightforward.
A: 
$$\int_0^1 x \left(\int_{x^2}^{x} \frac{1}{y} e^y dy \right) dx$$

You can integrate by parts. Let $u(x)=\int_{x^2}^{x} \frac{1}{y} e^y dy$ and $dv=xdx$. Then we have,
$$du=\left(\frac{e^x}{x}-(2x)\frac{e^{x^2}}{x^2}\right) dx$$
$$=\left(\frac{e^x}{x}-2\frac{e^{x^2}}{x} \right) dx$$
We may let $v=\frac{x^2}{2}$. Then because $\int u dv=uv-\int v du$ we get that the value we want is,
$$\int_{0}^{1} \left(xe^{x^2}-\frac{1}{2}xe^{x} \right) dx$$
This is a simple exercise in further integration by parts and substitution. One may show this leads to,
$$=\frac{1}{2}(e-2)$$
