Define $F(x) = \sum_{n\geq 1} f_{n}x^n$ and $G(x) = \sum_{n\geq 1} g_{n}x^n$. Then the Hadamard product of $F$ and $G$ is
$$H(x):=(F*G)(x) = \sum_{n\geq 1} f_{n}g_{n}x^n.$$
The author of Riesz equivalent of Riemann hypothesis and Hadamard product claims that
$$H(x) = \frac{1}{2\pi}\int_{0}^{2\pi} F(\sqrt {x} e^{it})G(\sqrt {x} e^{-it}) \mathrm{d}t.$$
However, no reference/proof of this identity was given. So, does anyone know where I can find the proof/reference of this identity ?
E.C. Titchmarsh, The theory of functions, Oxford University Press
Section 4.6 Hadamard multiplication theorem, p.158
Two derivations from operstional calculus:
For
$$A(x) = \sum_{n \geq 0} a_n x^n$$
and
$$\widetilde{A}(x) = \sum_{n \geq 0} a_n \frac{x^n}{n!} = e^{a.x}$$
with $(a.)^n = a_n$, the Hadamard product is given by
$$\sum_{n \geq 0} a_n x^n \frac{D_{x=0}^n}{n!} G(x)= \sum_{n\geq 0} a_ng_n x^n $$
with $d/dx= D_x$,
or more concisely,
$$\widetilde{A}(:xD_{x=0}:)G(x)= \exp(a.:xD_{x=0}:)G(x)=G(a.x)= (A*G)(x)$$
with $:xD_x:^n = x^nD_x^n$, by definition, a notational convenience.
The derivatives may be coded as the Cauchy contour integrals
$$g_n = \frac{D^n_{z=0}}{n!}G(z) = \frac{1}{2\pi i} \oint_{|z|<\epsilon} \frac{G(z)}{z^{n+1}}dz$$
where $\epsilon$ is less than the radii of the circles of convergence of the two series.
So, with appropriate changes of variables,
$$H(x)= (F*G)(x)$$
$$ = \frac{1}{2\pi i} \sum_{n \geq 0} f_n x^n \oint_{|z|<\epsilon} \frac{G(z)}{z^{n+1}}dz.$$
$$= \frac{1}{2\pi i} \oint_{|z|<\epsilon} \frac{F(\frac{x}{z})G(z)}{z}dz$$
$$= \frac{1}{2\pi i} \oint_{|z|<\alpha} \frac{F(\frac{\sqrt{x}}{z})G(z\sqrt{x})}{z}dz$$
$$= \frac{1}{2\pi} \int_0^{2\pi} F(\sqrt{x}\alpha^{-1}e^{-it})G(\sqrt{x}\alpha e^{it})dt$$
$$= \frac{1}{2\pi} \int_{0}^{2\pi} F(\sqrt{x}e^{-it})G(\sqrt{x}e^{it})dt,$$
assuming both series reps are convergent for $\alpha=1$. The last real integral is convergent for all functions bounded in the segment of integration.
For some discussion of the validity of these formulas, see "Hadamard grade of power series" by Allouche and France.
Alternatively, note (cf. this MSE answer)
$$\exp(txD_x)f(x)=f(e^t x).$$
Then
$$\frac{1}{2\pi} \int_{-\pi}^{\pi} F(ue^{-it})G(ve^{it})dt$$
$$= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-ituD_u}e^{itvD_v} dt F(u)G(v)$$
$$= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-it(uD_u-vD_v)} dt F(u)G(v)$$
$$=\frac{sin[\pi(uD_u-vD_v)]}{\pi(uD_u-vD_v)}F(u)G(v)$$
$$= \sum_{j,k \geq 0} sinc(\pi(j-k)) f_j g_k u^jv^k$$
$$= \sum_{k \geq 0} f_k g_k (uv)^k.$$
