First step, I reversed the order of integration (not sure if correct) of…
$$\int_0^1\int_{e^y}^e \frac x {\ln x} \, dx\,dy$$
to…
$$\int_0^e \int_{\ln x}^1 \frac x {\ln x} \, dy \, dx$$
I then began to evaluate…
$$\int_0^e \Big[\frac{xy}{\ln x}\Big]_{y=\ln x}^{y=1} \, dx$$
Plugging in gives me…
$$\int_0^e \left(\frac x {\ln x} – x\right) \, dx$$
Now I am left with an expression that can only be solved using integration by parts. I tried using integration by parts but I keep getting stuck. Online resources are not helping me. Can someone show me how to evaluate the rest step-by-step?
Or if my steps are incorrect please tell me.
Best Answer
\begin{align} \int_0^1 \left( \int_{e^y}^e \frac x {\ln x} \, dx \right) dy = \int_1^e \left( \int_0^{\ln x} \frac x {\ln x} \, dy \right) dx \end{align} You have $e^y \le x\le e.$ But that implies $y\le\ln x.$ So you cannot have $y$ going from $\ln x$ up to $1.$ The pair $(x,y)$ lies on one side of the curve $x = e^y$ or $\ln x = y,$ and that side is where $e^y\le x$ or $y \le \ln x.$
Draw the graph, showing the region over which you are integrating, and see where it's boundaries are, and that will make it clear what the bounds must be.
Notice that in $\displaystyle \int_0^{\ln x} \frac x {\ln x} \, dy,$ the function $\dfrac x {\ln x}$ does not change as $y$ goes from $0$ to $\ln x.$ That makes evaluation of the integral very easy.