1. **State the problem:** We want to find the inverse Laplace transform of the function given by the infinite sum $$F(s) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} s^{m+1}.$$ 2. **Rewrite the sum:** Recall that $\Gamma(m+1) = m!$. So the function is $$F(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^{m+1}.$$ 3. **Express the inverse Laplace transform:** The inverse Laplace transform $f(t)$ is given by $$f(t) = \mathcal{L}^{-1}\{F(s)\}(t) = \mathcal{L}^{-1}\left\{ \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^{m+1} \right\}(t).$$ 4. **Use linearity of inverse Laplace transform:** We can interchange sum and inverse transform (assuming convergence): $$f(t) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} \mathcal{L}^{-1}\{ s^{m+1} \}(t).$$ 5. **Find inverse Laplace transform of $s^{m+1}$:** Recall that for $n > -1$, $$\mathcal{L}\{ t^n \}(s) = \frac{\Gamma(n+1)}{s^{n+1}} \implies \mathcal{L}^{-1}\{ s^{-k} \}(t) = \frac{t^{k-1}}{\Gamma(k)}.$$ Here, $s^{m+1} = s^{-( -m -1)}$, so the exponent is positive, which is not a standard Laplace transform form. Usually, Laplace transforms involve negative powers of $s$. 6. **Rewrite $s^{m+1}$ as $s^{-( -m -1)}$:** This corresponds to the Laplace transform of a fractional derivative or distribution, which is complicated. Instead, consider the function as a formal power series in $s$. 7. **Alternative approach:** Consider the function $$G(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^{m}$$ so that $$F(s) = s G(s).$$ 8. **Inverse Laplace transform of $F(s)$:** Using the property $$\mathcal{L}^{-1}\{ s G(s) \}(t) = \frac{d}{dt} \mathcal{L}^{-1}\{ G(s) \}(t).$$ 9. **Find inverse Laplace transform of $G(s)$:** $$G(s) = \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^{m} = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} s^{m}.$$ 10. **Recognize $G(s)$ as a generalized Mittag-Leffler function:** The generalized Mittag-Leffler function is defined as $$E_{\alpha, \beta}(z) = \sum_{m=0}^\infty \frac{z^m}{\Gamma(\alpha m + \beta)}.$$ Here, the numerator is $m!$ instead of 1, so rewrite $m! = \Gamma(m+1)$ and consider the function $$H(z) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\Gamma(\alpha m + \beta)} z^m.$$ 11. **Express $\Gamma(m+1)$ as an integral:** Using the integral representation $$\Gamma(m+1) = \int_0^\infty x^m e^{-x} dx,$$ we can write $$H(z) = \sum_{m=0}^\infty \frac{z^m}{\Gamma(\alpha m + \beta)} \int_0^\infty x^m e^{-x} dx = \int_0^\infty e^{-x} \sum_{m=0}^\infty \frac{(x z)^m}{\Gamma(\alpha m + \beta)} dx.$$ 12. **Recognize the inner sum as $E_{\alpha, \beta}(x z)$:** So $$H(z) = \int_0^\infty e^{-x} E_{\alpha, \beta}(x z) dx.$$ 13. **Therefore, the inverse Laplace transform of $G(s)$ is related to the inverse Laplace transform of $H(s)$:** This integral representation is complicated, but the key insight is that $$f(t) = \frac{d}{dt} \mathcal{L}^{-1}\{ G(s) \}(t) = \frac{d}{dt} g(t),$$ where $$g(t) = \mathcal{L}^{-1}\{ G(s) \}(t).$$ 14. **Summary:** The inverse Laplace transform of the original function is $$f(t) = \frac{d}{dt} \left( \mathcal{L}^{-1} \left\{ \sum_{m=0}^\infty \frac{m!}{\Gamma(\alpha m + \beta)} s^m \right\} (t) \right).$$ 15. **Final note:** Without additional constraints on $\alpha, \beta$, or $s$, the inverse Laplace transform involves generalized Mittag-Leffler functions and their integral representations. The problem is advanced and typically requires special function theory for explicit closed forms. **Answer:** The inverse Laplace transform is $$f(t) = \frac{d}{dt} \left( \int_0^\infty e^{-x} E_{\alpha, \beta}(x t) dx \right),$$ where $E_{\alpha, \beta}$ is the generalized Mittag-Leffler function.