# 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 $$\Big[ \frac { 0 } { 2^n }, \frac { 1 } { 2^n }, \frac { 1 } { 2^n }, \frac { 2 } { 2^n }, \frac { 2 } { 2^n }, \ldots , \frac { 2^{n} - 1 } { 2^n }, \frac { 2^{n} - 1} { 2^n }, \frac { 2^{n} } { 2^n } \Big]$$.

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