Levy Khinchine formula for Gamma distribution

I am searching for the Levy-Khinchin formula for the Gamma distribution $$\Gamma(\lambda, \beta)$$. My attempt is following:

For the characteristic function I get

$$\varphi_{\Gamma(\lambda,\beta)}(t)=\Big(\frac{\beta}{\beta -it}\Big)^\lambda$$ Therefore

$$[\varphi_{\Gamma(\frac{\lambda}{n},\beta)}(t)]^n=\Big[\Big(\frac{\beta}{\beta -it}\Big)^\frac{\lambda}{n}\Big]^n=\varphi_{\Gamma(\lambda,\beta)}(t)$$

and the gamma distribution in infinitely divisible. Now the idea is to find a triplet $$(b,\sigma,\nu)$$ for the Levy-Khinchine formula. I have a hint to use $$\nu$$ which has a density regarding to the lebesgue measure as follows

$$\frac{\lambda}{x}e^{-\beta x}1_{[0,\infty)}(x)$$

So the Levy Kinchin formula is the following:

$$\psi(t)=itb-\frac{t^2\sigma^2}{2}+\int_{\mathbb{R}}\Big(e^{itx}-1-itx\cdot1_{\{|x|<1\}}(x)\Big)\frac{\lambda}{x}e^{-\beta x}1_{[0,\infty)}(x) dx$$

Now I take derivates of $$\log\circ\varphi_{\Gamma(\lambda,\beta)}$$ and $$\psi$$ to find $$b$$:

$$\psi'(t)=ib-t\sigma^2+\frac{\lambda i}{\beta-it}+\lambda i\Big(\frac{1}{\beta}e^{-\beta}-\frac{1}{\beta}\Big)$$

$$(\log\circ\varphi_{\Gamma(\lambda,\beta)})'(t)=\frac{\lambda i}{\beta-it}$$

So I take $$\sigma=0$$ and $$b=-\lambda \Big(\frac{1}{\beta}e^{-\beta}-\frac{1}{\beta}\Big)$$ to get $$\psi'(t)=(\log\circ\varphi_{\Gamma(\lambda,\beta)})'(t)$$. We also have $$\psi(0)=\varphi_{\Gamma(\lambda,\beta)}(0)=0.$$

Now I get

\begin{align} \varphi_{\Gamma(\lambda,\beta)}(t)&=\exp(\log(\varphi_{\Gamma(\lambda,\beta)}(t)))\\ &=\exp(\log(\varphi_{\Gamma(\lambda,\beta)}(t)-\log(\varphi_{\Gamma(\lambda,\beta)}(0)))\\ &=\exp(\int_0^t \psi'(s)ds)\text{, what we just calculated}\\ &=\exp(\psi(t)-\psi(0))\\ &=\exp(\psi(t)) \end{align}

Is this the right way?