evaluate $\displaystyle \int_{0}^{1} \int_{e^y}^{e}$ $\displaystyle \frac{x} {lnx} dydx$
Well $\displaystyle \int\frac{x}{e^{x}}$
Let $\displaystyle u=ln(x)\Rightarrow{e^u=x}$
So $\displaystyle dx=e^{u}$
So we would have $\displaystyle \int\frac{e^{2u}}{u}du$
Which has no general solution but
$\displaystyle e^{2u}=\sum_{n=0}^{\infty}\frac{2^nu^n}{n!}\Righta rrow{\frac{e^{2u}}{u}=\sum_{n=1}^{\infty}\frac{2^n u^{n-1}}{n!}}$
so $\displaystyle \int\frac{e^{2u}}{u}du=\sum_{n=0}^{\infty}\frac{2^ nu^n}{n\cdot{n!}}+C$
This double integral is improper?
My calculator agrees this integral tends towards ∞....best disregard my answer just in case I am wrong..because Mr. F seems to see something I dont
First of all, for the double integral to make any sense it should be written as
$\displaystyle \int_{0}^{1} \int_{e^y}^{e}$ $\displaystyle \frac{x} {lnx} d{\color{red}x} \, d{\color{red}y}$.
So you're integrating with respect to x first and then with respect to y. It can't be done in this order .... the x-integral can't be done easily, if at all (I haven't bothered checking).
So the order of integration has to be reversed.
The region of the xy-plane you're integrating over is defined by the curves $\displaystyle x = e^y \Rightarrow y = \ln x, ~ x = e, ~ y = 0$ and $\displaystyle y = 1$. You get these from the given integral terminals.
Draw the graphs of the curves and shade in this region. Then you should see that when you reverse the order of integration the double integral becomes:
$\displaystyle \int_{x = 1}^{x = e} \int_{y = 0}^{y = \ln x} \frac{x} {\ln x} \, d{\color{red}y} \, d{\color{red}x}$.
Integrating first with respect to y removes all the difficulties.
I get $\displaystyle \frac{e^2 - 1}{2}$ as the answer.