1. **Problem statement:** Find the inverse Laplace transform of the function given by the infinite series $$F(s) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} s^{m+1}.$$ 2. **Rewrite the series:** Note that $\Gamma(m+1) = m!$. So the series becomes $$F(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^{m+1} = s \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m.$$ 3. **Consider the inverse Laplace transform:** The inverse Laplace transform $f(t)$ satisfies $$f(t) = \mathcal{L}^{-1}\{F(s)\}(t) = \mathcal{L}^{-1}\left\{ s \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m \right\}(t).$$ 4. **Use linearity and known transforms:** The factor $s$ corresponds to differentiation in time domain: $$\mathcal{L}^{-1}\{s G(s)\} = \frac{d}{dt} g(t),$$ where $g(t) = \mathcal{L}^{-1}\{G(s)\}$. Let $$G(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m.$$ 5. **Find $g(t)$:** The inverse Laplace transform of $s^m$ is $$\mathcal{L}^{-1}\{s^m\} = \frac{t^{-m-1}}{\Gamma(-m)}$$ which is not defined for integer $m \geq 0$. Instead, consider the series as a formal power series in $s$ and interpret the inverse Laplace transform termwise: $$\mathcal{L}^{-1}\{s^m\} = \frac{t^{m}}{\Gamma(m+1)}$$ for $m \geq 0$ if we consider the Laplace transform of $t^m$ is $\frac{\Gamma(m+1)}{s^{m+1}}$ (note the inverse relationship). However, since powers of $s$ are positive, this is not a standard Laplace transform form. 6. **Alternative approach:** Rewrite $F(s)$ as $$F(s) = s \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m = s \cdot H(s),$$ where $$H(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m.$$ 7. **Recognize $H(s)$ as a generalized Mittag-Leffler function:** The Mittag-Leffler function is defined as $$E_{\alpha, \beta}(z) = \sum_{m=0}^\infty \frac{z^m}{\Gamma(\alpha m + \beta)}.$$ 8. **Relate $H(s)$ to $E_{\alpha, \beta}(z)$:** Since $m! = \Gamma(m+1)$, $$H(s) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} s^m,$$ which is not exactly the Mittag-Leffler function but can be expressed as $$H(s) = \sum_{m=0}^\infty m! \cdot \frac{s^m}{\Gamma(\alpha m + \beta)}.$$ 9. **Use the Laplace transform of the Mittag-Leffler function:** The Laplace transform of $$t^{\beta - 1} E_{\alpha, \beta}(\lambda t^\alpha)$$ is $$\mathcal{L}\{t^{\beta - 1} E_{\alpha, \beta}(\lambda t^\alpha)\}(s) = \frac{s^{\alpha - \beta}}{s^\alpha - \lambda}$$ for suitable parameters. However, our series is more complicated due to the $m!$ factor. 10. **Conclusion:** The inverse Laplace transform of the given series does not correspond to a standard closed-form function due to the factorial in numerator and Gamma function in denominator. It can be expressed formally as $$f(t) = \frac{d}{dt} \left( \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} \mathcal{L}^{-1}\{s^m\}(t) \right).$$ Since $\mathcal{L}^{-1}\{s^m\}$ is not standard for positive powers, the problem likely requires additional context or constraints for a closed form. **Final answer:** The inverse Laplace transform of $$F(s) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} s^{m+1}$$ is $$f(t) = \frac{d}{dt} \left( \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} \delta^{(m)}(t) \right),$$ where $\delta^{(m)}(t)$ is the $m$-th derivative of the Dirac delta function, representing a distributional solution rather than a classical function. This reflects the complexity and suggests the inverse Laplace transform is a generalized function involving derivatives of delta functions.