Proof of a Bromwich integral formula

Question

I am trying to prove that:

$$\frac{1}{2\pi i }\int_{\alpha-i\infty}^{\alpha+i\infty}\frac{(\log s)^{n}}{s}e^{sx}ds=(-1)^{n}\frac{d^{n}}{dz^{n}}\frac{x^{z}}{\Gamma(1+z)}\left.\begin{matrix} \\ \\ \end{matrix}\right|_{z=0}$$ Where $\alpha>0$.

Answer 1

We use the Laplace transform formula: $$(-1)^{n}\int_{0}^{\infty}\frac{d^{n}}{dz^{n}}\frac{x^{z}}{\Gamma(1+z)}e^{-sx}dx\left.\begin{matrix}\\ \\ \end{matrix}\right|_{z=0}=(-1)^{n}\frac{d^{n}}{dz^{n}}\frac{1}{\Gamma(1+z)}\int_{0}^{\infty}x^{z}e^{-sx}dx\left.\begin{matrix}\\ \\ \end{matrix}\right|_{z=0}$$ $$=(-1)^{n}\frac{d^{n}}{dz^{n}}\frac{1}{\Gamma(1+z)}\frac{\Gamma(1+z)}{s^{z+1}}\left.\begin{matrix}\\ \\ \end{matrix}\right|_{z=0}=\frac{(\log s)^{n}}{s}$$ And thus,by Bromwich integral formula: $$\frac{1}{2\pi i }\int_{\alpha-i\infty}^{\alpha+i\infty}\frac{(\log s)^{n}}{s}e^{sx}ds=(-1)^{n}\frac{d^{n}}{dz^{n}}\frac{x^{z}}{\Gamma(1+z)}\left.\begin{matrix}\\ \\ \end{matrix}\right|_{z=0}$$