Consider the integral
$$\int_0^\infty \frac{e^{-\lambda s}}{(1+is)^{1/2}}\,{\rm d}s$$
where $\lambda\to\infty$ is a positive scalar. My lecturer stated that the integral can be re-written in the form:
$$\int_0^\infty e^{-\lambda s}\sum_{n=0}^\infty \frac{(-is)^n\Gamma\left(n+\frac12\right)}{n!\Gamma\left(\frac12\right)}\,{\rm d}s$$
but I haven't ever met the Gamma function so I was wondering if someone could show me the manipulation in order to get it into this form.
By the binomial theorem we have that
$$(1+z)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}z^n$$
using this we find
$$(1+is)^{-1/2} = \sum_{n=0}^\infty \binom{-1/2}{n}(is)^n=\sum_{n=0}^\infty\frac{(-1/2)(-3/2)\cdots(-1/2-n+1)}{n!}(is)^n$$
How do I proceed? Also, his solution has $(-is)^n$ whereas mine is only $(is)^n$. How do you fix this?
This is simply an application of the binomial series to $(1+is)^{-1/2}$, which states that $$(1+z)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} z^k,$$ where $$\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots (\alpha-k+1)}{k!}$$ for non-integer $\alpha$. Then observe that $\Gamma(n+1/2)/\Gamma(1/2)$ is simply $(n-1/2)(n-3/2)\cdots(1/2)$.