Prove that when $\lambda$ is large enough

Question

\begin{align} f(z)= \frac{1}{2\pi j }\int_{\mathcal{C}}\frac{1}{ \lambda}\frac{\mathrm{\Gamma}(c+s)}{\mathrm{\Gamma}(c+1)}\frac{\mathrm{\Gamma}(b+1)\lambda^{s}}{\mathrm{\Gamma}(b+s)}z^{-s}ds=\frac{1}{2\pi j}\int_{\mathcal{C}}\frac{\mathrm{\Gamma}\left(\Phi-1+s\right)}{\mathrm{\Gamma}\left(\Phi\right)} \Psi^{-s+1}z^{-s}ds, \forall z \in \mathbb{R}^{*}_{+}, s\in \mathbb{C} \end{align}

Answer 1

The condition $g(z, s, \lambda) \sim h(z,s)$ as $\lambda\to\infty$ for each $z, s$, is equivalent to $$\frac{g(z, s, \lambda)}{h(z,s)}\to 1 \text{ as } \lambda \to \infty \text{ for each } z, s$$ and this is not strong enough to ensure the required result. However if instead we had $$\frac{g(z, s, \lambda)}{h(z,s)}\to 1 \text{ uniformly in $s$ as } \lambda \to \infty \text{ for each } z$$ it would work.

As a counterexample for the first, consider $D(s, \lambda)$ defined by

$D(s, \lambda)=|\lambda|$ for $-\frac{1}{\lambda}<s<\frac{1}{\lambda}$ and $D(s,\lambda)=0$ otherwise. Note that

$$\int_{0}^{\infty}D\left(s,\lambda \right)ds=1 \text{ for each $\lambda$, and that}$$ $$D(s,\lambda)\to0 \text{ as }\lambda \to\infty \text{ for each $s\ne 0$ .}$$

If we define $g(z,s,\lambda)=h(z,s)+D(s,\lambda)$, then we see that

$$g(z, s, \lambda) \sim h(z,s)\text{ as }\lambda\to\infty \text{ for each $z$ and $s$,}$$ since when $\lambda$ is large enough ($\lambda>|\frac{1}{s}|$), we have $D(s,\lambda)=0$. Also

$$f(z) = \int_{0}^{\infty}g\left(z,s,\lambda \right)ds = \int_{0}^{\infty}h((z,s)+D(s,\lambda) ds$$ $$= \int_{0}^{\infty}h((z,s) ds+1\ne\int_{0}^{\infty}h((z,s) ds$$

The result works with uniform convergence. We have $$ 1-\delta < \frac{g(z, s, \lambda)}{h(z,s)}<1+\delta \text{ whenever $\lambda>L(z,\delta)$, so (taking $h$ to be positive)}$$ $$ (1-\delta).h(z,s) < g(z, s, \lambda)<(1+\delta).h(z,s) \text{ whenever $\lambda>L$.}$$ It follows that $$ \frac{1}{1+\delta}g(z, s, \lambda)<h(z,s)< \frac{1}{1-\delta}g(z, s, \lambda) \text{ whenever $\lambda>L$. Integrating gives}$$ $$ \frac{1}{1+\delta}.f(z)<\int^{\infty}_{0}h(z,s)ds < \frac{1}{1-\delta}.f(z) \text{ and as this holds for every $\delta>0$ }$$ $$ f(z)=\int^{\infty}_{0}h(z,s)ds \text{ as required }$$