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The question is : if \begin{equation} \mathcal{M}[F(x)]=f(p) \end{equation}

find \begin{equation} \mathcal{M}[ln(x)\cdot x^3\cdot \frac{d^2}{dx^2} F(2x^3)] \end{equation}

atleast where can i start??

m7waath
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1 Answers1

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I will do this little exercise for you, based on 11.2.1.4 of this book , let our holomorphy strip be $(a,b)$, $\{s : a < Re(s) < b\}$

$\Large{\mathcal{M}[ F(x^3) ; s] = \frac{1}{3} f(\frac{s}{3})}$ where $s \in (3a, 3b)$

$\Large{\mathcal{M}[ F(2x^3) ; s] = \frac{1}{3 \cdot 2^{\frac{s}{3}}}f(\frac{s}{3})}$ where $s \in (3a, 3b)$

$\Large{\mathcal{M}[\partial_x^2 F(2x^3);s]}= \frac{(\frac{s}{3}-2)(\frac{s}{3}-1)}{3 \cdot 2^\frac{s}{3}}f(\frac{s}{3}-2)$ where $s \in (3a+2, 3b+2)$

$\Large{ \mathcal{M}[\frac{2x^3}{2} \partial_x^2 F(2x^3);s]}= \frac{(\frac{s}{3}-2)(\frac{s}{3}-1)}{2 \cdot 3 \cdot 2^\frac{s}{3}}f(\frac{s}{3}-2+1)$ where $s \in (3a+2-1, 3b+2-1)$

$\Large{ \mathcal{M}[\{\frac{1}{3}\ln(2x^3) - \frac{\ln(x)}{3} \} x^3 \partial_x^2 F(2x^3);s]}= \partial_s \lbrace\frac{(\frac{s}{3}-2)(\frac{s}{3}-1)}{2 \cdot 3 \cdot 2^\frac{s}{3}}f(\frac{s}{3}-2+1)\rbrace - \frac{\ln(2)(\frac{s}{3}-2)(\frac{s}{3}-1)}{3 \cdot 2 \cdot 3 \cdot 2^\frac{s}{3}}f(\frac{s}{3}-2+1)$ where $s \in (3a+1, 3b+1)$

I think you can leave the answer as is.

Note that multiplication $\cdot x^3$ is commutative with $\cdot \ln(x)$ whereas $\partial_x$ isn't.

Lada Dudnikova
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    The first identity should be $\mathcal M[F(a x^p)] = f(s/p)/(p a^{s/p})$. On the subsequent steps, you should apply the substitution $s = s + a$ to the whole expression, not just to the $f$ part: $$\frac d {ds} \left{ (s - 1) (s - 2) \left( \frac 1 {3 \cdot 2^{s/3}} f {\left( \frac s 3 \right)} \bigg\rvert_{s = s - 2} \right) \bigg\rvert_{s = s + 3} \right}.$$ – Maxim Apr 19 '19 at 13:36
  • @Maxim edited, could you check whether I did the line with differentiation correctly? – Lada Dudnikova Apr 19 '19 at 19:52
  • Not quite. On the previous step, we have $\mathcal M[G(x)] = g(s)$; then $\mathcal M[G''(x)] = (s - 1) (s - 2) g(s - 2)$, and now we recall that $g(s) = f(s/3)/(3 \cdot 2^{s/3})$. – Maxim Apr 19 '19 at 20:37