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This question already has an answer here:

Sophomore's dream states that: $$ \int_0^1x^{-x}dx=\sum_{n=1}^\infty n^{-n} $$ and $$ \int_0^1x^{x}dx=-\sum_{n=1}^\infty(-n)^{-n} $$ A friend of mine noticed that numerically: $$ \int_0^1\int_0^1(xy)^{xy}dxdy\approx 0.7834... \approx \int_0^1x^{x}dx $$ Are these two integrals really equal?

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marked as duplicate by Martin R, Community May 21 at 11:27

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    $\begingroup$ Why are the sum over $k$ when you have $n^{-n}$? Also you have an additional minus sign in the second equation. $\endgroup$ – user10354138 May 21 at 10:04
  • $\begingroup$ @user10354138 Opss a typo, thanks. Fixed $\endgroup$ – Picaud Vincent May 21 at 10:06
  • $\begingroup$ @MartinR I was not aware of it :-(. Hopefully my answer/computation is different. $\endgroup$ – Picaud Vincent May 21 at 11:20
  • $\begingroup$ @PicaudVincent: One can find many duplicates with Approach0 (see also math.meta.stackexchange.com/q/24978/42969) $\endgroup$ – Martin R May 21 at 11:24
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    $\begingroup$ @MartinR not aware of that too. Thanks. I am going to use it next time to avoid duplicates $\endgroup$ – Picaud Vincent May 21 at 11:27
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Change variable from $(x,y)$ to $(z = xy, y)$, we have $$\begin{align} \int_0^1 \int_0^1 (xy)^{xy} dx dy = &\int_0^1 \int_0^x \frac{z^z}{y} dz dy = \int_0^1 \int_x^1 \frac{z^2}{y} dy dz\\ = & \int_0^1 (- \log z) z^z dz = \int_0^1 z^z - (1+\log z) z^z dz\\ = &\int_0^1 z^z - (z^z)' dz = \int_0^1 z^z dz - [ z^z ]_0^1\\ = &\int_0^1 z^z dz \end{align} $$

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  • $\begingroup$ that's shorter! Thanks. What I wrote was inspired by the 1D approach you can find in wikipedia $\endgroup$ – Picaud Vincent May 21 at 10:24
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Yes they are equal! The trick is to use the binomial theorem to separate the integrals:

\begin{align} \int_0^1\int_0^1(xy)^{xy}dxdy &= \int_0^1\int_0^1\exp(xy\log{xy})dxdy \\ &= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy\log{xy})^ndxdy \\ &= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy)^n(\log{x}+\log{y})^ndxdy \\ &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(\int_0^1 x^n(\log{x})^kdx\right)\left(\int_0^1 y^n(\log{y})^{(n-k)}dy\right) \end{align}

Then using this result: $$ \int_0^1u^n\log{u}^m du=-m!(-\frac{1}{1+n})^{1+m} $$ we get: \begin{align} \int_0^1\int_0^1(xy)^{xy}dxdy &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(k!(-\frac{1}{1+n})^{1+k}\right)\left((n-k)!(-\frac{1}{1+n})^{1+n-k}\right) \\ &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nn!(-\frac{1}{1+n})^{2+n} \\ &= \sum_{n=0}^\infty (1+n)(-\frac{1}{1+n})^{2+n} \\ &= \sum_{n=0}^\infty (-1)^{2+n}(\frac{1}{1+n})^{1+n} \\ &= \sum_{n=1}^\infty -(\frac{-1}{n})^{n} \\ & = \int_0^1x^xdx \end{align}

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