# Much integrals. such definite. wow

**Calculus**Level 5

\[ { \huge u_0 = \int \limits_{ 0 } ^{ 1 } x \, dx , ~~~~~~~~~~ u_1 = \int \limits^{ \int \limits_{ 1/2 } ^{ 1 } x \, dx } _{ \int \limits_{ 0 } ^{ 1/2 } x \, dx } x \,dx , ~~~~~~~~ u_2 = \int \limits_{ \color{ red } { \int \limits_{\color{blue}{\int \limits _{0} ^{1/4} x \, dx}} ^{\color{blue}{\int \limits_{1/4} ^{2/4} x \, dx}} x \, dx}} ^{\color{red}{\int \limits_{\color{blue}{\int \limits_{2/4} ^{3/4} x \, dx}} ^{\color{blue}{\int \limits_{3/4} ^{1} x \, dx}} x \, dx}} x \,dx }\]

\( \{ u_{ i } \} _{ i = 0 } ^{ \infty } \) is a sequence of real numbers. The first few terms are as described above.

If \(P = \displaystyle \prod _{ n = 0 } ^{ 2015 } \frac{ 1 } { 4 u_{ n } }\), find \( \lfloor \log_{2} \log_{ 2 } P \rfloor \).

**Details and assumptions**

To clarify, in \(u_2\), the terms in blue are the limits of the red integrations, which in turn are the limits of the black integration.

The limits in \(u_0\) when read from bottom to top are \(\Big[ \frac{ 0 }{ 2^{ 0 } }, \frac{ 1 }{ 2^{ 0 } } \Big] \).

The limits in \(u_1\) when read from bottom to top are \(\Big[ \frac { 0 } { 2^1 }, \frac { 1 } { 2^1 }, \frac { 1 } { 2^1 }, \frac { 2 } { 2^1 } \Big] \).

The limits in \(u_2\) when read from bottom to top are \(\Big[ \frac { 0 } { 2^2 }, \frac { 1 } { 2^2 }, \frac { 1 } { 2^2 }, \frac { 2 } { 2^2 }, \frac { 2 } { 2^2 }, \frac { 3 } { 2^2 }, \frac { 3 } { 2^2 }, \frac { 4 } { 2^2 } \Big] \).

In general, the limits in \(u_n\) when read from bottom to top are \(\Big[ \frac { 0 } { 2^n }, \frac { 1 } { 2^n }, \frac { 1 } { 2^n }, \frac { 2 } { 2^n }, \frac { 2 } { 2^n }, \ldots , \frac { 2^{n} - 1 } { 2^n }, \frac { 2^{n} - 1} { 2^n }, \frac { 2^{n} } { 2^n } \Big] \).