Home » Uncategorized

Category Archives: Uncategorized

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.

Read more

Multiple logarithmic integral

Let \zeta denote the Riemann zeta function. Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln \prod_{k=1}^{n} x_k \ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\]

Solution

Based on symmetries,

    \begin{align*} \mathcal{J} &=n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln x_{1}\ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &=-n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1}\ln x_{1}\sum_{i=1}^{ \infty}\frac{(x_{1}\dots x_{n})^i}{i}\, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\\ &=-n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n-1}}\int_{0}^{1}x_{1}^i\ln x_{1}dx_{1} \\ &=n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}} \end{align*}

Let \displaystyle \mathcal{S}_n = \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}}. It follows that

    \begin{align*} \mathcal{S}_n &= \sum_{i=1}^{\infty} \frac{1}{(i+1)^n} \left ( \frac{1}{i} - \frac{1}{i+1} \right ) \\ &= \mathcal{S}_{n-1} - \sum_{i=2}^{\infty} \frac{1}{i^{n+1}} \\ &= \mathcal{S}_{n-1} + 1 - \zeta (n +1) \end{align*}

Using the recursion we get that

    \[\mathcal{S}_n = n + 1 - \sum_{k=2}^{n+1} \zeta(k)\]

Thus,

    \[\mathcal{J} = n \left( n +1 - \sum_{k=2}^{n+1} \zeta(k) \right)\]

Read more

An integral equality

Let \psi^{(1)} denote the trigamma function and let f be integrable on (0, 1). It holds that

    \[\int_{0}^{1} f \left ( \left \{ \frac{1}{x} \right \} \right )\, \mathrm{d}x = \int_{0}^{1} f(x) \psi^{(1)} (1+x) \; \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \int_{0}^{1} f \left(\left\{\frac{1}{x}\right\}\right) \mathrm{d}x &= \sum_{k=1}^{\infty}\int_{1/(k+1)}^{1/k} f \left(\left\{ \frac{1}{x} \right\} \right) \mathrm{d}x \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left( \left\{ u \right\} \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f\left( u-k \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{0}^{1} f\left( v \right) \: \frac{\mathrm{d} v}{(v+k)^{2}} \\ &= \int_{0}^{1} f\left( v \right) \sum_{k=1}^{\infty} \frac{1}{(v+k)^{2}} \: \mathrm{d} v \\ &= \int_{0}^{1} f(v) \psi^{(1)}(v+1) \: \mathrm{d}v \end{align*}

where the interchange between the infinite sum and the integration is allowed by the uniform bound

    \[\left|\sum_{k=1}^{N} \frac{1}{(v+k)^{2}} \right| < \frac{\pi^{2}}{6} \quad , \quad N \geq 1, \, 0 \leq v \leq 1\]

The result follows.

Read more

 

Digamma series

Let \psi^{(0)} denote the digamma series. Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( \psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n} \right)\]

Solution

The series telescopes;

    \begin{align*} \mathcal{S} &=\sum_{n=1}^{\infty} \bigg(\psi^{(0)} \left(\frac{1+2n-1}{2} \right) -\psi^{(0)} \left( \frac{2n-1}{2} \right) - \\ &\quad \quad \quad -\frac{1}{2n-1}+\psi^{(0)} \left( \frac{1+2n}{2}\right)-  \psi^{(0)}\left(\frac{2n}{2} \right)-\frac{1}{2n} \bigg)\\ &=\sum_{n=1}^{\infty} \bigg( \psi^{(0)}\left( n \right)-\psi^{(0)} \left( n-\frac{1}{2} \right) -\frac{1}{2n-1}+ \\ &\quad \quad \quad +\psi^{(0)} \left( n+\frac{1}{2} \right)-\psi^{(0)} \left( n \right) -\frac{1}{2n} \bigg) \\ &=\sum_{n=1}^{\infty} \bigg( \psi^{(0)} \left(n+\frac{1}{2}\right) -\psi^{(0)} \left(n-\frac{1}{2}\right)- \\ &\quad \quad \quad -\frac{1}{2n-1}-\frac{1}{2n} \bigg)\\ &=\sum_{n=1}^{\infty} \left( \frac{2}{2n-1}-\frac{1}{2n-1}-\frac{1}{2n} \right) \\ &=\sum_{n=1}^{\infty} \left( \frac{1}{2n-1}-\frac{1}{2n} \right)\\ &=\ln 2 \end{align*}

Read more

Continuous and periodic

Let f be a continuous real-valued function on \mathbb{R} satisfying

    \[\left| f(x) \right| \leq \frac{1}{1+x^2} \quad  \forall x\]

Define a function F on \mathbb{R} by

    \[F(x) = \sum_{n=-\infty}^{\infty} f \left ( x + n \right )\]

  1. Prove that F is continuous and periodic with period 1.
  2. Prove that if G is continuous and periodic with period 1 then

        \[\int_{0}^{1} F(x)G(x) \, \mathrm{d}x = \int_{-\infty}^{\infty} f(x) G(x) \, \mathrm{d}x\]

Solution

  1. We note that

        \[g\left ( x + 1 \right ) = \sum_{n=-\infty}^{\infty} f \left ( n +x +1 \right ) = \sum_{n'=-\infty}^{\infty} f \left ( x + n \right ) = g(x)\]

  2. First of all G is bounded on [0, 1] and \sum \limits_{n=-N}^{N} f (x + n ) \rightarrow F(x) uniformly. Hence,

        \begin{align*} \int_{-\infty}^{\infty} f(x) G(x)\, \mathrm{d}x &= \sum_{n=-\infty}^{\infty} \int_{n}^{n+1} f(x) G(x) \, \mathrm{d}x \\ &= \sum_{n=-\infty}^{\infty} \int_{0}^{1} f(x + n) G(x)\, \mathrm{d}x \\ &= \int_{0}^{1} F(x) G(x) \, \mathrm{d}x \end{align*}

Read more

Donate to Tolaso Network