The Mellin transform of the function $h$, locally integrable on $(0,\infty)$, is defined by

$$M[h,z] = \int_0^\infty t^{z-1} h(t) dt \tag{1}$$

For some functions $h$ the above integral is not analytic anywhere, yet the Mellin transform can be defined using truncation and analytic continuation. My question is about finding a closed form of the Mellin transform in such a case.

As a specific example, let $n\geq 0$ be an integer, and $h=Y_n$ be the Bessel function of the second kind. Set

$$f(z) = - \frac{1}{\pi} 2^{z-1} \cos \frac{(z-n) \pi}{2} \Gamma\left(\frac{z+n}{2}\right) \Gamma\left(\frac{z-n}{2}\right),$$

where $\Gamma$ is the Gamma function. Clearly, $f$ is a meromorphic function in $\mathbb{C}$. From the tables of Mellin transforms we find

$$M[h; z] = f(z), \qquad n<\Re z < \frac{3}{2}.$$

In the case $n\geq 2$ this formula is invalid as the above strip of analyticity is an empty set. This means that the Mellin transform of $h$ cannot be defined using the original integral definition $(1)$. It is instead defined by truncations and analytic continuations

$$M[h; z] = M[h \chi_{(0,1)}; z] + M[h \chi_{(1,\infty)}; z] =: f_1(z) + f_2(z)$$

The first term is analytic in $\Re z > n$, the reason is that $h(t) \sim a_n t^{-n}$ as $t \to 0$ so that the integral in $(1)$ is convergent in this half-plane. The second term is analytic in $\Re z < \frac{3}{2}$ because of the asymptotics $h(t) \sim b_n t^{-\frac{1}{2}} e^{it}$ as $t \to + \infty$. It is known that both $f_1$ and $f_2$ can be continued analytically as meromorphic functions in $\mathbb{C}$, as a result $M[h; z]$ is also a meromorphic function.

I am wondering how can one find a closed form for $M[h; z]$ (expressed in terms of special functions). Is there a relation between $M[h; z]$ and $f(z)$ in the case $n \geq 2$?

(Upon looking through the tables of Mellin transforms, we find many other functions for which the given formula becomes invalid as described above for large enough $n$, e.g. $h = H_n^{(1)}, Y_n^2$ etc.)

The Mellin transform of the function $h$, locally integrable on $(0,\infty)$, is defined by

$$M[h,z] = \int_0^\infty t^{z-1} h(t) dt \tag{1}$$

For some functions $h$ the above integral is not convergent anywhere, yet the Mellin transform can be defined using truncation and analytic continuation. My question is about finding a closed form of the Mellin transform in such a case.

As a specific example, let $n\geq 0$ be an integer, and $h=Y_n$ be the Bessel function of the second kind. Set

$$f(z) = - \frac{1}{\pi} 2^{z-1} \cos \frac{(z-n) \pi}{2} \Gamma\left(\frac{z+n}{2}\right) \Gamma\left(\frac{z-n}{2}\right),$$

where $\Gamma$ is the Gamma function. Clearly, $f$ is a meromorphic function in $\mathbb{C}$. From the tables of Mellin transforms we find

$$M[h; z] = f(z), \qquad n<\Re z < \frac{3}{2}.$$

In the case $n\geq 2$ this formula is invalid as the above strip of analyticity is an empty set. This means that the Mellin transform of $h$ cannot be defined using the original integral definition $(1)$. It is instead defined by truncations and analytic continuations

$$M[h; z] = M[h \chi_{(0,1)}; z] + M[h \chi_{(1,\infty)}; z] =: f_1(z) + f_2(z)$$

The first term is analytic in $\Re z > n$, the reason is that $h(t) \sim a_n t^{-n}$ as $t \to 0$ so that the integral in $(1)$ is convergent in this half-plane. The second term is analytic in $\Re z < \frac{3}{2}$ because of the asymptotics $h(t) \sim b_n t^{-\frac{1}{2}} e^{it}$ as $t \to + \infty$. It is known that both $f_1$ and $f_2$ can be continued analytically as meromorphic functions in $\mathbb{C}$, as a result $M[h; z]$ is also a meromorphic function.

I am wondering how can one find a closed form for $M[h; z]$ (expressed in terms of special functions). Is there a relation between $M[h; z]$ and $f(z)$ in the case $n \geq 2$?

(Upon looking through the tables of Mellin transforms, we find many other functions for which the given formula becomes invalid as described above for large enough $n$, e.g. $h = H_n^{(1)}, Y_n^2$ etc.)