1. **State the problem:** We want to find the inverse Laplace transform of the function given by the infinite sum: $$F(s) = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\left(\Gamma(\alpha m + \beta)\right)^{m+1}}$$ where $\Gamma$ is the Gamma function, and $\alpha, \beta$ are parameters. 2. **Analyze the sum:** The Gamma function $\Gamma(m+1)$ is equivalent to $m!$ for integer $m$. So the numerator is $m!$. The denominator is $\left(\Gamma(\alpha m + \beta)\right)^{m+1}$, which is a power of the Gamma function evaluated at a linear function of $m$. 3. **Consider the Laplace transform context:** Typically, the Laplace transform $\mathcal{L}\{f(t)\}(s)$ is an integral transform involving $e^{-st}$. The inverse Laplace transform recovers $f(t)$ from $F(s)$. 4. **Rewrite the sum for clarity:** $$F(s) = \sum_{m=0}^\infty \frac{m!}{\left(\Gamma(\alpha m + \beta)\right)^{m+1}}$$ Since the problem is to find the inverse Laplace transform of this sum, we need to identify if this sum corresponds to a known Laplace transform of a function or can be expressed in a closed form. 5. **Check for simplifications or known series:** Without additional context or values for $\alpha$ and $\beta$, the sum is quite complex. However, if $\alpha$ and $\beta$ are such that $\Gamma(\alpha m + \beta)$ simplifies or relates to factorials, we might express the sum in terms of known functions. 6. **Assuming $\alpha=1$ and $\beta=1$ for illustration:** Then $\Gamma(\alpha m + \beta) = \Gamma(m+1) = m!$. Substitute: $$F(s) = \sum_{m=0}^\infty \frac{m!}{(m!)^{m+1}} = \sum_{m=0}^\infty \frac{m!}{(m!)^{m+1}} = \sum_{m=0}^\infty \frac{1}{(m!)^m}$$ This is a convergent series but does not correspond directly to a standard Laplace transform. 7. **General approach to inverse Laplace transform:** If $F(s)$ is given explicitly, the inverse Laplace transform $f(t)$ is: $$f(t) = \mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} e^{st} F(s) ds$$ But here, $F(s)$ is a sum independent of $s$, so the problem might be misstated or missing $s$ dependence. 8. **Conclusion:** Since the sum does not depend on $s$, its inverse Laplace transform is a distribution concentrated at $t=0$ scaled by the sum value: $$f(t) = \left( \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\left(\Gamma(\alpha m + \beta)\right)^{m+1}} \right) \delta(t)$$ where $\delta(t)$ is the Dirac delta function. **Final answer:** The inverse Laplace transform of the given sum (which is independent of $s$) is a scaled delta function at zero: $$f(t) = C \delta(t) \quad \text{where} \quad C = \sum_{m=0}^\infty \frac{\Gamma(m+1)}{\left(\Gamma(\alpha m + \beta)\right)^{m+1}}$$