# Even stronger than Sophomore's dream [duplicate]

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?

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

• Why are the sum over $k$ when you have $n^{-n}$? Also you have an additional minus sign in the second equation. – user10354138 May 21 at 10:04
• @user10354138 Opss a typo, thanks. Fixed – Picaud Vincent May 21 at 10:06
• @MartinR I was not aware of it :-(. Hopefully my answer/computation is different. – Picaud Vincent May 21 at 11:20
• @PicaudVincent: One can find many duplicates with Approach0 (see also math.meta.stackexchange.com/q/24978/42969) – Martin R May 21 at 11:24
• @MartinR not aware of that too. Thanks. I am going to use it next time to avoid duplicates – Picaud Vincent May 21 at 11:27

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}
\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}