I was trying to compute the 3rd integral, marked $z=re^{i\frac{\pi}{2p}}$ for $\oint{e^{i\lambda z^p}\, dz}$. I started with $z=re^{i\frac{\pi}{2p}}$ followed by $t=\lambda r^p$ :
$$\begin{align}\int_{\Gamma}{e^{i\lambda z^p}\, dz}&=e^{i\frac{\pi}{2p}}\int_{0}^{\infty}{e^{-\lambda r^p}\, dr}\\&=p\lambda^{1/p}e^{i\frac{\pi}{2p}}\int_{0}^{\infty}{e^{-t}\, t^{1-1/p}\, dt}\\&=p\lambda^{1/p}e^{i\frac{\pi}{2p}}\Gamma\left(2-\frac{1}{p}\right)\end{align}$$
Trying with some numbers like $\lambda=1$ and $p=2$ seem like the solution is ok. The answer which this (page 2/3) document gives is $$e^{i\frac{\pi}{2p}}\lambda^{-1/p}\Gamma\left(\frac{1}{p}+1\right)$$
Are these really the same result or was the $\lambda=1$ and $p=2$ case just a coincidence?
What you seem to want is the computation of $$\int_0^\infty e^{i\lambda x^p} dx$$
For this you consider the path $\Gamma(R)$ which is the concatenation of the paths $\Gamma_1(R)$ on the real line, $\Gamma_2 (R)$ the arc part, and $\Gamma_3 (R)$ the line with the angle. You have $\int_{\Gamma_2(R)} e^{i \lambda z^p} \to 0$ when $R \to \infty$ which gives you:
$$A =\int_0^\infty e^{i\lambda x^p} dx = -\int_{\Gamma_3(\infty)} e^{i\lambda z^p} = \int_0^\infty \gamma'(x) e^{i\lambda \gamma(x)^p} dx $$ with $\gamma(x) = xe^{i\frac{\pi}{2p}}$
So
$$A = e^{i\frac{\pi}{2p}} \int_0^\infty e^{i^2\lambda x^p} dx = e^{i\frac{\pi}{2p}} \int_0^\infty e^{-\lambda x^p} dx $$ Take now $u = \lambda x^p$, i.e. $x = \lambda^{-1/p} u^{1/p}$ so $dx = \lambda^{-1/p} u^{1/p -1}\frac{du}{p}$ (this is where you made a mistake) You get $$A = e^{i\frac{\pi}{2p}}\frac{\lambda^{-1/p}}{p} \int_0^\infty u^{1/p -1}e^{-u} du = e^{i\frac{\pi}{2p}}\frac{\lambda^{-1/p}}{p} \Gamma(1/p)$$
But $x \Gamma(x) = \Gamma(x+1)$ so $$A = e^{i\frac{\pi}{2p}}\lambda^{-1/p} \Gamma(1+ 1/p)$$