If $1\leq \alpha$ show show that the gamma density has a maximum at $\frac{\alpha - 1}{\lambda} $ So using this form of the gamma density function:
$$g(t) = \frac{\lambda^{\alpha}t^{\alpha -1}e^{-\lambda t}}{\int_0^\infty t^{\alpha - 1} e^{-t} \, dt} $$
I would like to maximize this. Now i was thinking of trying to differentiate with respect to $t$, but that is becoming an issue because if i remember correctly the denominator is not a close form when the bound goes to infiniti
 A: You're trying to maximize $t\mapsto t^{\alpha-1} e^{-\alpha t}$ on the interval $0\le t <\infty$.  That is the same as maximizing the logarithm of that function, since the logarithmic function is increasing, and taking the logarithm makes it a bit simpler:
$$
\log\left( t^{\alpha-1} e^{-\alpha t} \right) = (\alpha-1)\log t - \alpha t.
$$
The derivative of this with respect to $t$ is
$$
\frac{\alpha-1} t - \alpha = \alpha\frac{1-(1/\alpha) - t} t.
$$
Since the denominator is positive, the question now is only when the numerator is positive, negative, and zero.  The numerator $1-(1/\alpha)-t$ is positive if $t<1-1/\alpha$ and negative if $t>1-1/\alpha$, and zero of $t=1-1/\alpha$.  Hence the original function increases on $[0,1-1/\alpha]$ and decreases on $[1-1/\alpha,\infty)$, reaching its peak at $1-1/\alpha$.
All this is true if $\alpha>1$; if $\alpha<1$ then the function decreases on $[0,\infty)$ and has its maximum at $t=0$.
A: The denominator is a constant as you are integrating it over $t$. So it is same as maximizing only the numerator.
And $\frac{d(t^{\alpha -1}e^{-\alpha t})}{dx}=(\alpha -1)t^{\alpha -2}e^{-\lambda t}-\lambda t^{\alpha -1}e^{-\lambda t}$.
Equating this to $0$ you get $t=0, ( \alpha -1)/\lambda$
Check the double derivative at these points to get the maximum.
