Definite parametric integral

Let 0<a<b. Evaluate the integral

    \[\mathcal{J} = \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x\]

Solution

The key substitution is x \mapsto \frac{ab}{u}. Applying it we see that

    \begin{align*} \sqrt{abx+ x^3} &\overset{x \mapsto ab/u}{=\! =\! =\! =\! =\!} \sqrt{\frac{a^2b^2}{u} +\frac{a^3b^3}{u^3} } \\ &=\sqrt{\frac{a^2b^2u^2}{u^3} + \frac{a^3b^3}{u^3}} \\ &=\sqrt{\frac{a^2b^2 \left ( u^2+ab \right )}{u^2 \cdot u}} \\ &=\frac{ab}{u} \sqrt{\frac{u^2+ab}{u}} \end{align*}

Thus,

    \begin{align*} \mathcal{J} &= \int_{a}^{b} \frac{e^{x/a}-e^{b/x}}{\sqrt{abx+ x^3}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{x=ab/u}{=\! =\! =\! =\! =\! =\!} ab\int_{a}^{b} \frac{1}{u^2} \cdot \left ( e^{b/u} - e^{u/a} \right ) \cdot \frac{u}{ab} \cdot \frac{\sqrt{u}}{\sqrt{u^2+ab}} \, \mathrm{d}u \\ &=\int_{a}^{b} \frac{e^{b/u}-e^{u/a}}{\sqrt{abu + u^3}} \, \mathrm{d}u \\ &= - \int_{a}^{b} \frac{e^{u/a}-e^{b/u}}{\sqrt{abu+u^3}} \, \mathrm{d}u \\ &= -\mathcal{J} \end{align*}

Thus , \mathcal{J}=0.

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Power of matrix

Let \displaystyle A=\begin{pmatrix} -2 & 4 &3 \\ 0 & 0 & 0\\ -1 &5 &2 \end{pmatrix}. Prove that A^{593}-2A^{15}+A=0.

Solution

The characteristic polynomial of A is p(x)=x-x^3. This in return means A=A^3 and A^3=A^5. Thus,

    \begin{align*} A^{593} -2 A^{15} +A &= A^{591} \cdot A^2 - 2 \left ( A^3 \right )^5 + A\\ &=\left ( A^3 \right )^{197} \cdot A^2 - 2 A^5 + A \\ &= A^{197} \cdot A^2 - 2 A^3 + A \\ &=A^{199} - 2 A +A\\ &=A^{198} \cdot A- A\\ &=\left ( A^3 \right )^{66} \cdot A - A \\ &=A^{66} \cdot A - A\\ &= \left ( A^3 \right )^{22} \cdot A - A\\ &= A^{22} \cdot A -A \\ &= A^{23} - A\\ &= \left ( A^3 \right )^{7} \cdot A^2 -A\\ &= A^7 \cdot A^2 - A\\ &= A^9 - A\\ &=\left ( A^3 \right )^3 - A\\ &=A^3 -A\\ &=A -A\\ &=\mathbb{O} \end{align*}

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Gamma infinite product

Prove that

    \[\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}} = \frac{8\sqrt{\pi}}{e^2}\]

Solution

Converting the product to a sum and using duplication formula for the gamma function and telescoping,

\displaystyle \sum_{n=1}^{N}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-N}\right)\ln\left(2\sqrt{\pi}\right)-2N\ln2+\frac{2}{2^{N+1}}\ln\Gamma(2^{N+1})

Using Stirling formula

    \[\frac{1}{N}\ln\Gamma(N)=\ln N-1+\mathcal{O}\left(\frac{\ln N}{N}\right)\quad \text{as}\;\; N\rightarrow\infty\]

we get that

    \[\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}\]

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Upper bound of max product

Let z_1, z_2 , \dots, z_n \in \mathbb{C} be the roots of the polynomial

    \[f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{C}[x]\]

Prove that:

    \[\prod_{k = 1}^{n} \max (1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Solution

Suppose the roots of polynomial f are z_1, z_2, \dots, z_n where

    \[|z_1| \geq |z_2| \geq \cdots \geq |z_m| > 1 \geq |z_{m+1}| \geq \cdots \geq |z_n|\]

Let g(z)=z^nf \left(\frac{1}{z}\right) = 1 + a_{n-1}z + \cdots + a_0z^n. Then, the \{1/z_k\}_{1\leq k \leq m} are the zeros of g in the disk |z| \le r = 1-\epsilon < 1 where \epsilon is chosen such that g(re^{i\theta}) \neq 0 for \theta \in [0, 2\pi].

Jensen’s inequality implies that

    \begin{align*} \log|r^m z_1 \cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d} \theta & \Leftrightarrow \\ |r^m z_1\cdots z_m| = \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) &\Leftrightarrow \\ \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta \end{align*}

Applying Cauchy – Schwartz yields,

    \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta &\leq \frac{1}{2\pi}\left(\int_0^{2\pi}\,\mathrm{d}\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,\mathrm{d}\theta\right)^{1/2} \\ &= \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2} \end{align*}

Therefore,

    \[\left|r^m z_1\cdots z_m\right| \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Letting r\rightarrow 1 and \epsilon \rightarrow 0 we get the result.

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Harmonic limit

Let \mathcal{H}_n denote the n – th harmonic number. Evaluate the limit

    \[\ell =\lim_{n \rightarrow +\infty} \left ( \frac{1}{\ln n} \sum_{k=1}^{n} \frac{\mathcal{H}_k}{k} - \frac{\ln n}{2} \right )\]

Solution

Lemma: It holds that \displaystyle \sum_{k=1}^{n} \frac{\mathcal{H}_k}{k} = \frac{\mathcal{H}_n^2+\mathcal{H}_n^{(2)}}{2}.

Proof: We have successively:

    \begin{align*} 2\sum_{k=1}^n \frac{\mathcal{H}_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}\\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=k}^n \frac{1}{jk}\\ &= \sum_{k=1}^n \sum_{j=1}^n \frac{1}{jk} + \sum_{k=1}^n \frac{1}{k^2} \\ &= \left(\sum_{k=1}^n \frac{1}{k} \right)^2 + \sum_{k=1}^n \frac{1}{k^2} \\ &= \mathcal{H}_n^2+ \mathcal{H}^{(2)}_n \\ \end{align*}

Thus,

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

 

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