On the dubious function

The dubious function \Delta:\mathbb{N} \rightarrow \mathbb{N} is defined as follows : \Delta(1)=1 and

    \[\mathbf{\Delta(n) = \sum_{\substack{\mathbf{d \mid n } \\ \mathbf{d \neq n} }} \Delta (d)} \quad \text{for} \;\; n>1\]

Evaluate the sum

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\Delta\left ( 15^n \right )}{15^n}\]

Definite logarithmic integral

Let 0<\alpha<\beta. Evaluate the integral:

    \[\mathcal{J}=\int_\alpha^\beta \frac{\ln x}{(x+\alpha)(x+\beta)}\, \mathrm{d}x\]

Solution

We have successively:

    \begin{align*} \mathcal{J} &= \int_{\alpha}^{\beta} \frac{\ln x}{\left ( x+\alpha \right )\left ( x+\beta \right )}\, \mathrm{d}x \\ &\!\!\!\!\!\!\!\overset{x \mapsto \alpha \beta/x}{=\! =\! =\! =\! =\!=\!} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta- \ln x}{\left ( x + \alpha \right ) \left ( x + \beta \right )} \, \mathrm{d}x\\ &=\frac{1}{2} \int_{\alpha}^{\beta} \frac{\ln \alpha \beta}{\left ( x+\alpha \right )\left ( x + \beta \right )} \, \mathrm{d}x \\ &=\frac{\ln \alpha \beta}{\beta-\alpha} \ln \left ( \frac{(\alpha+\beta)^2}{4\alpha \beta} \right ) \end{align*}

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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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