Evaluate for a double integral

Question

Evaluate $$\int_{0}^{1} \int_{e^y}^{e} \frac{x}{\ln x}\ dx\ dy$$

I'm new to double integral. Can anyone guide me for this question? Thanks in advance

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}\int_{\expo{y}}^{\expo{}}{x \over \ln\pars{x}}\,\dd x\,\dd y & \,\,\,\stackrel{x\ =\ \exp\pars{t}}{=}\,\,\, \int_{0}^{1}\int_{y}^{1}{\expo{t} \over t}\,\expo{t}\,\dd t\,\dd y = \int_{0}^{1}{\expo{2t} \over t}\int_{0}^{t}\,\dd y\,\dd t = \int_{0}^{1}\expo{2t}\,\dd t \\[5mm] & = \bbx{\mrm{e}^{2} - 1 \over 2} \approx 3.1945 \end{align}

Answer 2

The idea of multiple integrals is to treat them as iterated integrals. For example, the integral provided can be rewritten as $$\int_{y=0}^{y=1}\left(\int_{x = e^y}^{x=e}\frac{x}{\ln x} dx\right)dy$$ It should be clear that we must evaluate the inner integral first, and the inner integral must resolve to a function of $y$. Assuming you could integrate the inner integral (let's call it $F(x)$), the double integral becomes $$\int_0^1(F(e)-F(e^y))dy$$ All that remains is to evaluate this final integral.

Answer 3

Hint. This double integral is improper because the integrand is not bounded along the line $x=1$. Note that $$\int_{y=0}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy =\lim_{\epsilon\to 0} \int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy.$$ By changing the order of integration (Fubini's Theorem), we get $$\int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy= \int_{x=e^{\epsilon}}^{e} \left(\int_{y=\epsilon}^{\ln(x)} \frac{x}{\ln x} dy\right) dx=\int_{x=e^{\epsilon}}^{e} \frac{x}{\ln x}(\ln(x)-\epsilon) dx.$$ Can you take it from here?

P.S. In order to avoid the limit you should use a version of Fubini's Theorem for improper integrals.

Answer 4

This integral requires a change in the order of integration to be computed. You will see that the area over which we are integrating is equivalent to $1\leq y\leq e; 0\leq x \leq \ln y$. The best way to check this is simply to graph it. As a result, we can rewrite this integral to be: $$\int_1^e\int_0^{\ln x}\frac x{\ln x}\,dydx$$ Evaluating the inside integral: $$\int_1^e\frac x{\ln x}\left(\ln x - 0\right)dx$$ And we can then use single variable methods to get a final answer of $\frac 12\left(e^2-1\right)$.