If $$\lim_{t \to 0} \left(\int_{0}^{1} (3x+5)^t \, dx\right)^{\frac{1}{t}} = \frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$$, then $$\alpha$$ is equal to ________

This is in the form $$1^\infty$$. Let $$L$$ be the limit. Take $$\ln L$$:

$$\ln L = \lim_{t\to 0} \frac{\ln \int_{0}^{1} (3x + 5)^t \, dx}{t}$$

Using L'Hôpital's Rule (differentiating w.r.t $$t$$):

$$\ln L = \lim_{t\to 0} \frac{\frac{d}{dt} \int_{0}^{1} e^{t\ln(3x+5)} \, dx}{\int_{0}^{1} (3x+5)^t \, dx} = \frac{\int_{0}^{1} \ln(3x+5) \, dx}{1}$$

Evaluate $$\int_{0}^{1} \ln(3x+5) \, dx$$ using substitution $$u = 3x+5, du = 3dx$$:

$$\frac{1}{3} \int_{5}^{8} \ln u \, du = \frac{1}{3} [u \ln u - u]_5^8 = \frac{1}{3} (8 \ln 8 - 8 - 5 \ln 5 + 5) = \ln\left(\frac{8^8}{5^5}\right)^{1/3} - 1$$

 $$L = e^{\ln(8^8/5^5)^{1/3} - 1} = \frac{1}{e} \cdot \frac{8^{8/3}}{5^{5/3}} = \frac{1}{e} \cdot \frac{8^2 \cdot 8^{2/3}}{5^1 \cdot 5^{2/3}} = \frac{64}{5e} \left(\frac{8}{5}\right)^{2/3}$$

Comparing gives $$\alpha = \mathbf{64}$$.