Is the gamma function expressible as a proper integral of elementary functions? You're also allowed to compose it with however many elementary functions. But strictly no limits.
[edit] So far the answers move the limit inside the integral. Is that still elementary?
On
You can also do this: $$\begin{align*}\Gamma(t) &= \int_{x=0}^\infty x^{t-1} e^{-x} \, dx \\ &= \int_{x=0}^1 x^{t-1} e^{-x} \, dx + \int_{x=1}^\infty x^{t-1} e^{-x} \, dx \\ &= \int_{x=0}^1 x^{t-1} e^{-x} \, dx + \int_{u=1}^0 (u^{-1})^{t-1} e^{-1/u} (-u^{-2}) \, du \\ &= \int_{x=0}^1 x^{t-1} e^{-x} + x^{-t-1} e^{-1/x} \, dx, \end{align*}$$ and the function $$f(t;x) = x^{t-1} e^{-x} + x^{-t-1} e^{-1/x}$$ is bounded on $x \in (0,1)$ for $t > 1$.
Yes, although you probably won't like this much: consider $$ \int_0^1 2x \left(-2\log{x}\right)^{n-1} \, dx. $$ Because $x (\log{(1/x)})^n \to 0$ as $x \to 0$, the integrand is continuous and bounded on $(0,1]$. Hence this is a proper integral. Now, set $x^2 = e^{-t}$, so $$ 2x \, dx = - e^{-t} \, dt, $$ and $ -2\log{x} = -2\log{(e^{-t/2})} = t $. Then the integral becomes $$ \int_0^{\infty} t^{n-1} e^{-t} \, dt = \Gamma(n). $$