How can you show using Watson's lemma, that for some infinitely differentiable function $K(s)$ and $ kt \gg 1$ that
$$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$
Does one say, let us change variables to $u=t-s$
$$\int_0^tK(s) e^{-k(t-s)} \text{d}s = \int_t^0 K(t-u) e^{-ku} (-\text{d} u) \\ = \int_0^t K(t-u) e^{-ku} \text{d}u \approx \frac{K(t) \Gamma(1)}{k} $$ By Watson's lemma as $ kt \gg 1$ and $ k \gg 1$ (the $n=0$ term in the summation of Watson's lemma).
This clearly doesn't work, would someone be able to point out where I have gone wrong here and how I can produce the desired result?
EDIT
Something I have tried is
$$\int_0^t K(t-u) e^{-ku} \text{d}u = \int_0^\varepsilon K(t-u) e^{-ku} \text{d}u + O(e^{k\varepsilon})$$
where $1/k\ll \varepsilon \ll 1$. Letting $z=ku$ gives $$\frac{1}{k}\int_0^{\varepsilon k} K\left(t-\frac{z}{k}\right)e^{-z}\text{d}z \sim \frac{K(t)}{k}\int_0^{\infty}e^{-z}\text{d}z$$ as $\varepsilon k \gg 1$ and by making the approximation $K\left(t-\frac{z}{k}\right)\sim K(t)$ as $k\rightarrow\infty$.
Therefore as $z =k(t-s)$ $$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$
Question update
The above method gives the correct result, where then in my original attempt have I made a mistake and specifically what mistake was it?
Note, the definition for Watson's lemma which I used originally is found here https://en.wikipedia.org/wiki/Watson%27s_lemma
$$\int_0^t K(t-u) e^{-ku} \text{d}u = \int_0^\varepsilon K(t-u) e^{-ku} \text{d}u + O(e^{k\varepsilon})$$
where $1/k\ll \varepsilon \ll 1$. Letting $z=ku$ gives $$\frac{1}{k}\int_0^{\varepsilon k} K\left(t-\frac{z}{k}\right)e^{-z}\text{d}z \sim \frac{K(t)}{k}\int_0^{\infty}e^{-z}\text{d}z$$ as $\varepsilon k \gg 1$ and by making the approximation $K\left(t-\frac{z}{k}\right)\sim K(t)$ as $k\rightarrow\infty$.
Therefore as $z =k(t-s)$ $$\int_0^tK(s) e^{-k(t-s)} \text{d}s\approx K(t)\int_{-\infty}^t e^{-k(t-s)} \text{d}s$$
This is equivalent to the initial work you did you massive cretin.
Many thanks to Antonio Vargas in pointing this out.