Parametric improper integral

Let \alpha, \beta>0 such that \alpha \neq \beta. Prove that

    \[\int_{0}^{\infty} \frac{1-x^2}{\left ( \alpha x +\beta \right )^2} \frac{\ln\left ( 1+x \right )}{\left ( \beta x + \alpha \right )^2} \, \mathrm{d}x = \frac{1}{\alpha \beta \left ( \alpha^2 - \beta^2 \right )} \ln \frac{\beta}{\alpha}\]

Improper Gamma integral

Let \Gamma denote the Euler’s Gamma function. Prove that

    \[\int_{-\infty}^{\infty}\frac {\mathrm{d}x} {\Gamma(\alpha+x)\Gamma(\beta-x)}=\frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}\]

where \mathfrak{Re}(\alpha + \beta) >1.

Arithmotheoretic sum

Evaluate the limit:

    \[\ell= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m\]

Solution

    \begin{align*} \ell &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m \\ &= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} \left ( n - m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \left ( n^2 - \sum_{m=1}^{n} m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=1 - \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{m=1}^{n} \frac{m}{n} \left \lfloor \frac{n}{m} \right \rfloor \\ &= 1 - \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x \end{align*}

However,

    \begin{align*} \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x &\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{1}^{\infty} \frac{\left \lfloor u \right \rfloor}{u^3} \, \mathrm{d}x \\ &=\sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n}{x^3} \, \mathrm{d}x \\ &= \frac{1}{2}\sum_{n=1}^{\infty} n \left ( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{n}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{(n+1)^2} \right )\\ &=\frac{1}{2}\cancelto{1}{\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right )}+ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} \\ &=\frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \frac{1}{n^2} \\ &=\frac{1}{2} + \frac{1}{2} \left ( \frac{\pi^2}{6} - 1 \right ) \\ &=\frac{\pi^2}{12} \end{align*}

Hence

    \[\ell = 1 - \frac{\pi^2}{12}\]

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Series of Bessel function

Let J_n denote the Bessel function of the first kind. Prove that

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]

Solution

The Jacobi – Anger expansion tells us that

(1)   \begin{equation*} e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}  \end{equation*}

Hence by Parseval’s Theorem it follows that

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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Proof of “Fermat’s last theorem”

Let x,y,z,n\in \mathbb{N}^* and n \geq z. Prove that the equation

    \[x^n+y^n=z^n\]

has no solution.

Solution

Without loss of generality , assume that x<y. If x^n+y^n=z^n held , then it would be z^n > y^n thus z^n \geq (y+1)^n. It follows from Bernoulli’s inequality that,

    \begin{align*} z^n &=x^n +y^n\\ &< 2y^n \\ &\leq \left(1 + \frac {n}{z} \right) y^n \\ &\leq \left(1 + \frac {1}{z} \right)^ny^n\\ &< \left(1 + \frac {1}{y} \right)^ny^n \\ &= (y+1)^n \\ &\leq z^n \end{align*}

which is an obscurity. The result follows.

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