Substitution in integration 2

Question

It seems like I dont quite understand the third line in Proof.

Now $$\int_0^\infty e^{-st} t^{1/2} dt = s^{-1/2} \int_0^\infty e^{-u} u^{-1/2} du $$

$ u:= st$, $du/dt=s$ therefore $ dt= du/s $ I can't come up with the right term

since

$$\int_0^\infty e^{-st} t^{1/2} =\int_0^\infty e^{-u} \frac{t^{1/2}}{s} du$$

But $ u \ne t/s$

Would be nice to have some explaination. thank you. Greetings.

Answer 1

$u=st$, hence $t=u/s$:

$$ \int_0^\infty e^{-st} t^{1/2} dt =\int_0^\infty e^{-u} \frac{(u/s)^{1/2}}{s} du =s^{-3/2} \int_0^\infty e^{-u} u^{1/2} du =s^{-3/2}\frac{\sqrt{\pi}}{2}. $$

Answer 2

$$ \int_0^\infty e^{-st} t^{1/2} \, dt = s^{-3/2} \int_0^\infty e^{-st}(st)^{1/2} (s\,dt) = s^{-3/2} \int_0^\infty e^{-u} u^{1/2} \, du $$