We now give examples that include functions that do not belong to Hardy's class for which Ramanujan's master theorem is applicable. As Γ(*s*) Γ(1-*s*) = *π*/sin(*πs*), using Hardy's notation we have

\varphi \left(-s\right)=F\left(s;0\right)/\Gamma \left(1-s\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\pi /\text{sin}\left(\pi s\right)\right)\varphi \left(-s\right)=\Gamma \left(s\right)F\left(s;0\right),

(see (1.6)). Therefore, it is natural to introduce classes of functions based on the asymptotic representations of the product Γ(*s*)*F*(*s*; 0) as we have done in Section 4.

**Example (5.1)** Let

\Gamma \left(s\right)F\left(s;0\right):=\Gamma \left(s\right)\zeta \left(s\right)\phantom{\rule{1em}{0ex}}\left(0<\sigma <1\right).

(5.1)

Then, we note that *F*(*s*, 0): = *ζ*(*s*) (-∞ < *σ* < 1) is analytic and for 0<\theta \le {\theta}_{0}<\frac{\pi}{2} and (see [[9], pp. 135-136])

\left|\Gamma \left(s\right)F\left(s,0\right)\right|\le C{\left(2\pi \right)}^{\sigma}{\text{e}}^{-\left(\frac{\pi}{2}-\theta \right)\left|\tau \right|}\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1\right).

(5.2)

Therefore, F\left(s;0\right):=\zeta \left(s\right)\in {Q}_{\text{log}\left(1/2\pi \right)}\left(\frac{\pi}{2}-{\theta}_{0},1\right). Hence, according to the lemma (3.1), the IMT

F\left(0;x\right)=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\Gamma \left(s\right)\zeta \left(s\right){x}^{-s}\text{d}s\phantom{\rule{1em}{0ex}}\left(0<c<1\right),

(5.3)

is well defined. The integrand in (5.3) has simple poles at *s* = 0, -1, -2, .... Using Cauchy's residue theorem, we get

F\left(0;x\right)=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}\zeta \left(-n\right){x}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(0<x<2\pi \right),

(5.4)

that can be simplified further as (see [[6], p. 264-266] and [[9], p. 136])

F\left(0;x\right)=\frac{1}{{\text{e}}^{x}-1}-\frac{1}{x}=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}\zeta \left(-n\right){x}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(0<x<2\pi \right).

(5.5)

**Example (5.2)** Let

\Gamma \left(s\right)F\left(s;0\right):=\Gamma \left(s\right)\Gamma \left(1-s\right)=\frac{\pi}{\text{sin}\phantom{\rule{0.3em}{0ex}}\pi s}\phantom{\rule{1em}{0ex}}\left(0<\sigma <1\right).

(5.6)

Here the function *F*(*s*; 0) = Γ(1-*s*) (*σ* < 1) is analytic and

\left|\Gamma \left(s\right)F\left(s;0\right)\right|=\left|\Gamma \left(s\right)\Gamma \left(1-s\right)\right|\le C{\text{e}}^{-\pi \left|\tau \right|}\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1\right),

(5.7)

which shows that *F*(*s*; 0): = Γ(1-*s*) ∈ *Q*_{0}(*π*, 1). Therefore, according to the lemma (3.1), the IMT

F\left(0;x\right)=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\Gamma \left(s\right)\Gamma \left(1-s\right){x}^{-s}\text{d}s=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\frac{\pi}{\text{sin}\phantom{\rule{0.3em}{0ex}}\pi s}{x}^{-s}\text{d}s\phantom{\rule{1em}{0ex}}\left(x>0,\phantom{\rule{0.3em}{0ex}}0<{\sigma}_{1}\le c\le {\sigma}_{2}<1\right),

(5.8)

is well defined. Using the Cauchy's residue theorem we find

F\left(0;x\right)=\frac{1}{1+x}=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}}{n!}\Gamma \left(1+n\right){x}^{n}=\sum _{n=0}^{\infty}{\left(-1\right)}^{n}{x}^{n}\phantom{\rule{1em}{0ex}}\left(0<x<1\right),

(5.9)

It is worth pointing out that Ramanujan's master theorem stated as

\underset{0}{\overset{\infty}{\int}}{x}^{s-1}\left(\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}\Gamma \left(1+n\right)}{n!}{x}^{n}\right)\text{d}x=\Gamma \left(s\right)\Gamma \left(1-s\right)=\frac{\pi}{\text{sin}\phantom{\rule{0.3em}{0ex}}\pi s}\phantom{\rule{1em}{0ex}}\left(0<\sigma <1\right),

(5.10)

is not meaningful, as the series in (5.9) and (5.10) is divergent if *x* > 1. However, the IMT (5.8) is well defined that has the series representation (5.9). The series representation in (5.10) must be replaced by the analytic continuation on the LHS of (5.9).

**Example (5.3)** We now give an example of a function where Ramanujan's master theorem is applicable but the function does not belong to Hardy's class of functions.

Consider F\left(s;0\right):=\text{sin}\phantom{\rule{0.3em}{0ex}}\left(\frac{\pi s}{2}\right). Then we note that

\left|\Gamma \left(s\right)\text{sin}\left(\frac{\pi s}{2}\right)\right|=O\left({\left|\tau \right|}^{\sigma -\frac{1}{2}}\right)\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1/2\right).

(5.11)

Hence, according to the above lemma (3.1), the line integral

F\left(0;x\right):=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\Gamma \left(s\right)\text{sin}\phantom{\rule{0.3em}{0ex}}\left(\frac{\pi s}{2}\right){x}^{-s}\text{d}s\phantom{\rule{1em}{0ex}}\left(x>0,\phantom{\rule{0.3em}{0ex}}0<{c}_{1}\le c\le {c}_{2}<1/2\right),

(5.12)

is well defined. From Cauchy's residue theorem, the integral (5.11) can be evaluated by taking the sum of the residues at the simple poles at *s* = -1, -3, -5, ... leading to a closed form *f*(*x*) = sin *x*. Hence, we have the Mellin transform representation

M\left[\text{sin}\phantom{\rule{0.3em}{0ex}}x;s\right]:=\underset{0}{\overset{\infty}{\int}}\frac{\text{sin}\phantom{\rule{0.3em}{0ex}}x}{{x}^{1-s}}\text{d}x=\Gamma \left(s\right)\text{sin}\phantom{\rule{0.3em}{0ex}}\left(\frac{\pi s}{2}\right)\phantom{\rule{1em}{0ex}}\left(0<\sigma <1/2\right).

(5.13)

We note that *F*(*s*; 0) ∈*Q*_{∞}(1, 1/2). Following Jordan's lemma, the integral (5.13) remains convergent for 0 < *σ* < 1 which shows that (3.9) is valid in the strip 0 < *σ* < 1. Note that in this example we have *A* = *π* and \varphi \left(-s\right)=F\left(s;0\right)/\Gamma \left(1-s\right)=\text{sin}\phantom{\rule{0.3em}{0ex}}\left(\frac{\pi s}{2}\right)/\Gamma \left(1-s\right) which shows that *ϕ* does not belong to Hardy's class of functions.

**Example (5.4)** We give a second example of a function where Ramanujan's master theorem is applicable but the function does not belong to Hardy's class of functions. The applicability of the Ramanujan's master theorem is assured by the lemma (3.1). Consider F\left(s;0\right)=\frac{1}{\Gamma \left(1-s\right)} (*σ* < 1/2). Then (see [[7], p. 6(1.45)])

\left|\frac{\Gamma \left(s\right)}{\Gamma \left(1-s\right)}\right|=O\left({\left|\tau \right|}^{2\left(\sigma -\frac{1}{2}\right)}\right)\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1/2\right).

(5.14)

Hence, according to the above lemma, the integral

\begin{array}{c}F\left(0;x\right):=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\frac{\Gamma \left(s\right)}{\Gamma \left(1-s\right)}{x}^{-s}\text{d}s\\ \left(x>0,\phantom{\rule{0.3em}{0ex}}0<{c}_{1}\le c\le {c}_{2}<1/2\right),\end{array}

(5.15)

is well defined. From Cauchy's residue theorem, the integral (5.15) can be evaluated by taking the sum of the residues at the simple poles at *s* = 0, -1, -2, ... leading to

F\left(0;x\right)=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}{x}^{n}}{{\left(n!\right)}^{2}},

(5.16)

which has the closed form F\left(0;x\right)={J}_{0}\left(2\sqrt{x}\right). Again we have *A* = *π* and

\varphi \left(-s\right)=F\left(s;0\right)/\Gamma \left(1-s\right)=1/{\left(\Gamma \left(1-s\right)\right)}^{2},

(5.17)

which shows that *ϕ* does not belong to Hardy's class of functions. However, *F*(s; 0) ∈*Q*_{∞}(2, 1/2).

**Example (5.5)** We give another example of a function involving ratio of the Riemann zeta functions where our lemma (3.1) assures the application of Ramanujan's master theorem but the function does not belong to Hardy's class of functions. Consider F\left(s;0\right):=\frac{\zeta \left(s\right)}{\Gamma \left(1-s\right)\zeta \left(1-s\right)} (*σ* < 1/2). Then (see [[5], p. 81])

\left|\frac{\Gamma \left(s\right)\zeta \left(s\right)}{\Gamma \left(1-s\right)\zeta \left(1-s\right)}\right|=O\left({\left|\tau \right|}^{\sigma -\frac{1}{2}}\right)\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1/2\right).

(5.18)

Hence, according to the above lemma, the integral

F\left(0;x\right):=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\frac{\Gamma \left(s\right)\zeta \left(s\right)}{\Gamma \left(1-s\right)\zeta \left(1-s\right)}{x}^{-s}\text{d}s\phantom{\rule{1em}{0ex}}\left(x>0,\phantom{\rule{0.3em}{0ex}}0<{c}_{1}\le c\le {c}_{2}<1/2\right),

(5.19)

is well defined. From Cauchy's residue theorem, the integral (5.16) can be evaluated by taking the sum of the residues at the simple poles at *s* = -1, -3, ... leading to

F\left(0;x\right)=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}\zeta \left(-n\right){x}^{n}}{{\left(n!\right)}^{2}\zeta \left(1+n\right)}.

(5.20)

The series (5.19) can be simplifies further by using Riemann functional equation (1.11) to give the closed form

F\left(0;x\right)=\sum _{n=0}^{\infty}\frac{{\left(-1\right)}^{n}\zeta \left(-n\right){x}^{n}}{{\left(n!\right)}^{2}\zeta \left(1+n\right)}=\frac{1}{\pi}\text{sin}\phantom{\rule{0.3em}{0ex}}\left(\frac{\pi x}{2}\right).

(5.21)

Once again we have

\varphi \left(-s\right)=F\left(s;0\right)/\Gamma \left(1-s\right)=\zeta \left(s\right)/\zeta \left(1-s\right){\left(\Gamma \left(1-s\right)\right)}^{2},

(5.22)

which shows that *ϕ* does not belong to Hardy's class of functions. However, *F*(s; 0) ∈*Q*_{∞}(1, 1/2).

**Example (5.6)** Consider

F\left(s;0\right):=\frac{\zeta \left(1-s\right)}{\Gamma \left(s\right)\zeta \left(s\right)}\phantom{\rule{1em}{0ex}}\left(\sigma <1/2\right).

(5.23)

Then we note that (see [[5], p.81])

\left|\Gamma \left(s\right)F\left(s;0\right)\right|=\left|\frac{\zeta \left(s\right)}{\zeta \left(1-s\right)}\right|=\left|\chi \left(s\right)\right|=O\left({\left|\tau \right|}^{\sigma -\frac{1}{2}}\right)\phantom{\rule{1em}{0ex}}\left(\left|\tau \right|\to \infty ,\phantom{\rule{0.3em}{0ex}}-\infty <{\sigma}_{1}\le \sigma \le {\sigma}_{2}<1/2\right).

(5.24)

Hence, according to the lemma (3.1), the IMT

F\left(0;x\right)=\frac{1}{2\pi i}\underset{c-i\infty}{\overset{c+i\infty}{\int}}\frac{\zeta \left(1-s\right)}{\zeta \left(s\right)}{x}^{-s}\text{d}s\phantom{\rule{1em}{0ex}}\left(0<c<1/2,\phantom{\rule{0.3em}{0ex}}x\ge 0\right),

(5.25)

is well defined. The integral (5.23) can be evaluated by Cauchy's residue theorem which has the closed form (see [[13], p. 91(3.3.6)]), *F*(0; *x*) = 2cos(*πx*). Once again the function

\varphi \left(-s\right)=F\left(s;0\right)/\Gamma \left(1-s\right)=\zeta \left(1-s\right)/\zeta \left(s\right)\Gamma \left(s\right)\Gamma \left(1-s\right),

(5.26)

which shows that *ϕ* does not belong to Hardy's class of functions. However, *F*(s; 0) ∈*Q*_{∞}(1, 1/2).