Moment generating function of continuous random variable

Question

I got a problem that said "find the Moment-generating function of $$f(t)=\frac{1}{4}e^{-t/4}\mathbb{I}_{(0,\infty)}(t)$$ And I solved it like this but I don't know if this' right $$\begin{align*}&M_{T}(y)=E\left [ e^{yt} \right ]\\&=\int_{0}^{\infty}e^{yt}\frac{1}{4}e^{-t/4}\,dt\\&=\frac{1}{4}\int_{0}^{\infty}e^{yt-t/4}\,dt \end{align*}$$ so $\lim_{t\rightarrow \infty}\frac{1}{4}\int_0^\infty e^{yt-t/4} \,dt$

Let $u=yt-\frac{t}{4}\Rightarrow du=(y-\frac{1}{4})dt \Rightarrow \frac{1}{y-\frac{1}{4}}\, du=dt$ $$\begin{align*}&\lim_{t\rightarrow \infty}\frac{1}{4}\int_0^{yt-\frac{t}{4}} e^u \frac{1}{y-\frac{1}{4}} \, du\\&=\lim_{t\rightarrow \infty}\frac{1}{4(y-\frac{1}{4})} \int_0^{yt-\frac{t}{4}} e^u \, du\\&=\lim_{t\rightarrow \infty}\frac{e^{yt-\frac{t}{4}}-1}{4y-1} \end{align*}$$ So $M_T(y)=\lim_{t\rightarrow \infty}\frac{e^{yt-\frac{t}{4}}-1}{4y-1}$

Answer 1

Almost finished. $(4y-1)^{-1}$ is a constant independent of $t$ and comes out of the limit. The limit does not exist if $\Re y>1/4$ and converges to $-1$ if $\Re y<1/4$ as $e^{-\infty}=0$, so we get $M_t(y)=(1-4y)^{-1}$ when $\Re y<1/4$.

In fact the distribution you are given is an exponential distribution with parameter $\lambda=1/4$. The MGF for this distribution with general paramater $\lambda$ can be found here.

Answer 2

I would rephrase it a little because you have a line starting with 'so' that is not going anywhere (the upper limit of the integral is still $\infty$ and the integrator 'dt' is missing). You can even do it without substitution: $$M_{T}(y)=E\left [ e^{yT} \right ]=\int_0^\infty e^{yt}\frac{1}{4}e^{-t/4} \, dt = \frac{1}{4}\int_0^\infty e^{(y-1/4)t}\,dt = \lim_{u \to \infty} \frac{1}{4} \int_0^u e^{(y-1/4)t} \, dt $$ $$=\lim_{u \to \infty} \frac{1}{4}\frac{1}{y-1/4} \left[ e^{(y-1/4)t}\right]_{t=0}^u=\lim_{u \to \infty} \frac{1}{4y-1} \left( e^{(y-1/4)u} - e^{(y-1/4)0} \right)$$ If $y-1/4 \geq 0$ the integral does not converge, else it simplifies to $1/(1-4y)$.