Prove that a function is a density : $f(x) = \exp ( r ( c+a) x - r e^{ax} ) $

Question

I have the following function :

$$ \forall x \in \mathbb R_+ , \qquad f(x) = \frac{ 1}{ R }\exp ( r ( c+a) x - r e^{ax} )$$

where $$ R := \frac 1 a r^{ - r ( 1 + \frac{c} {a} )} \overline{ \Gamma } \left( r, r \cdot ( 1 + \frac c a ) \right ) $$

and finally, the incomplete complementary gamma function :

$$\overline{ \Gamma } ( x, \theta ) = \int_x^{\infty} z^{\theta - 1 } e^{-z} dz$$

I'd like to show that :

$$ \int_{\mathbb R } f(x) dx = 1 $$

I find it particularly hard because of the fact that the incomplete gamma function does not have a closed form, and I haven't found for the exp part of $f$ an antiderivative.

What do you think?

------- EDIT:

Following @Michh's comment:

$$ \int \exp ( r ( c+a) x - r e^{ax} ) dx = \int u^{\frac{r (a + c) }{a} } \exp ( e^{-(r+a) u } ) du $$

there was indeed a mistake due to copy from paper:

$$ \int \exp ( r ( c+a) x - r e^{ax} ) dx = \int u^{\frac{r (a + c) }{a} } \exp ( e^{-r u } ) \frac 1 { a u } du $$

Answer 1

You have a mistake in the change of variable formula. Making the substitution $y = re^{ax}$ so that $x =\frac{1}{a} \log(y/r)$, we get $$\int_0^\infty \exp\left(r(c+a)x - re^{ax}\right) \, dx = \int_r^\infty \left(\frac{y}{r}\right)^{r(c+a)/a} e^{-y}\,\frac{dy}{ay} = \frac{1}{ar^{r(c+a)/a}}\Gamma\left(r,\frac{r(c+a)}{a}\right).$$