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I'm being asked to calculate $$I\triangleq\int_0^1\int_{e^{\large x}}^e{xe^y\over(\ln y)^2}\,dy\,dx\quad.$$

I got stuck on the indefinite inner one, $$J\triangleq\int{e^ydy\over(\ln y)^2}\quad.$$ At first, I tried substitution with $u=e^y$, $u=\ln y$ and $u=(\ln y)^2$, none of them useful. Then I looked up Wolfram Alpha and it says $J$ can't be written in terms of elementary functions.

I assume there's an analytical way to find $I$ without $J$, otherwise the answer is what Wolfram said and the estimate from the Double Integral Calculator that $I\approx6.21799$.

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2 Answers 2

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You just need to change limits. The region $0\leq x\leq 1$, $e^x\leq y \leq e$ can also be described as $1\leq y \leq e$, $0\leq x\leq \ln y$. Then $$ \int_0^1\int_{e^{\large x}}^e{xe^y\over(\ln y)^2}\,dy\,dx\quad=\int_1^e\int_0^{\ln y}\frac{xe^y}{(\ln y)^2}\,dx\,dy=\frac12\,\int_1^ee^y\,dy=\frac{e^e-e}2 $$

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  • $\begingroup$ Next time I'll know better than to not draw the freaking region. =D $\endgroup$
    – Luke
    May 7, 2013 at 4:50
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Note that $e^x < y$ iff $ x < \ln(y)$, and so the above can be rewritten as $$ \int_0^1 \int_{e^x}^e \frac{x e^y}{(\ln(y))^2}dydx = \int_1^e \int_0^{\ln(y)} \frac{x e^y}{(\ln(y))^2}dxdy = \int_1^e \frac{ e^y}{2(\ln(y))^2} (\ln(y))^2 dy $$

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