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Sophomore’s dream constant

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \int_0^1 (xy)^{xy} \,\mathrm{d}(x, y)\]

Solution

Let t=xy and s=y. The Jacobian is

    \[\frac{\partial (s, t)}{\partial (x, y)} = y \Rightarrow \left (\frac{\partial (s, t)}{\partial (x, y)} \right )^{-1} = \frac{1}{y} = \frac{1}{s}\]

Hence,

    \begin{align*} \mathcal{J} &= \iint \limits_{[0, 1]^2} \left ( xy \right )^{xy} \, \mathrm{d}(x, y) \\ &=\int_{0}^{1} \int_{0}^{s} \frac{t^t}{s} \, \mathrm{d}(t, s) \\ &= \int_{0}^{1} \int_{t}^{1} \frac{t^t}{s} \, \mathrm{d} ( s, t)\\ &= -\int_{0}^{1} t^t \log t \, \mathrm{d}t \end{align*}

However , since \displaystyle \int_{0}^{1} t^t \left ( 1 + \log t \right ) \, \mathrm{d}t =0 we conclude that

    \[\mathcal{J} = \int_0^1 t^t \, \mathrm{d}t = \mathcal{S}\]

where \mathcal{S} is Sophomore’s dream constant.

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Logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \ln^2 \left | \sqrt{x} - \sqrt{1-x} \right |\, \mathrm{d}x\]

An arcosine integral

Evaluate the integral

    \[\mathcal{J} = \int_{-1}^{1} \frac{\arccos x}{\sqrt{3x^4+2x^2+3}} \, \mathrm{d}x\]

A limit

For any nonnegative integer n, define

    \[\mathcal{J}_n = \int_{0}^{\pi/2} \frac{\sin^2 nt}{\sin t}\, \mathrm{d}t\]

Evaluate the limit \ell = \lim \limits_{n \rightarrow +\infty} \left( 2 \mathcal{J}_n - \ln n \right).

Solution

Lemma: Let n \in \mathbb{Z} and x \neq \kappa \pi. It holds that

    \[\sin x + \sin 3x + \cdots + \sin(2n-1)x = \frac{\sin^2 nx}{\sin x}\]

Proof: The LHS is just the imaginary part of

    \[\mathcal{S}=e^{ix}+e^{3ix}+\cdots+e^{(2n-1)ix}\]

This is a geometric progression , hence:

    \begin{align*} \mathcal{S} &=e^{ix}\frac{e^{2in x}-1}{e^{2ix}-1}\\ &=e^{inx}\frac{e^{inx}-e^{-inx}}{e^{i x}-e^{-ix}} \\ &=e^{inx}\frac{\sin nx}{\sin x}\\ &=(\cos nx+i\sin nx) \\ &=\frac{\sin nx}{\sin x} \end{align*}

The result follows. \blacksquare

Hence,

    \begin{align*} \mathcal{J}_n &= \int_{0}^{\pi/2} \frac{\sin^2 nt}{\sin t}\, \mathrm{d}t \\ &= \int_{0}^{\pi/2} \sum_{k=1}^{n} \sin \left ( 2k-1 \right )t \, \mathrm{d}t \\ &=\sum_{k=1}^{n} \int_{0}^{\pi/2} \sin \left ( 2k-1 \right )t \, \mathrm{d}t \\ &=\sum_{k=1}^{n} \frac{1}{2k-1} \\ &=\sum_{k=1}^{2n}\frac{1}{k}-\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k} \end{align*}

Now,

    \begin{align*} \ell &= \lim_{n \rightarrow +\infty} \left ( 2 \mathcal{J}_n - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} - \ln n \right ) \\ &=\lim_{n \rightarrow +\infty} \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - 2 \ln 2n + 2 \ln 2n - \sum_{k=1}^{n} \frac{1}{k} - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left [ \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - 2 \ln 2n \right ) + 2 \ln 2 - \left ( \sum_{k=1}^{n} \frac{1}{k} -\ln n \right ) \right ] \\ &= 2 \gamma + 2 \ln 2 - \gamma \\ &= \gamma + 2 \ln 2 \end{align*}

 

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Fractional integral

Let \{ \cdot \} denote the fractional part. Prove that

    \[\int_{0}^{\pi/2}\sin 2x \{\ln^{2n-1} \tan x \} \, \mathrm{d}x\]

for the different values of the integer number n.

Solution

Let \mathcal{J} denote the integral,

    \begin{align*} \mathcal{J} &= \int_{0}^{\pi/2} \sin 2x \left \{ \ln^{2n-1} \tan x \right \}\, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \sin 2 \left ( \frac{\pi}{2} - u\right ) \left \{ \ln^{2n-1} \tan \left ( \frac{\pi}{2} - u \right ) \right \}\, \mathrm{d}u \\ &= \int_{0}^{\pi/2} \sin 2u \left \{ \ln^{2n-1} \frac{1}{\tan u} \right \} \, \mathrm{d}u\\ &=\int_{0}^{\pi/2} \sin 2u \left \{ - \ln^{2n-1} \tan u \right \}\, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \left ( \left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} \right ) \, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \, \mathrm{d}u \\ &= \frac{1}{2} \end{align*}

since \left \{ x \right \} + \left \{ -x \right \} = 1 if x \notin \mathbb{Z} whereas \left \{ x \right \} + \left \{ -x \right \} = 0 if x \in \mathbb{Z}. Therefore,

    \[\left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} =1\]

except of a countable set \displaystyle A = \left\{ x_k \in \left ( 0, \frac{\pi}{2} \right ) \bigg| \ln^{2n-1} \tan x_k \in \mathbb{Z} \right\} whose measure is 0.

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