Evaluate $\int_{0}^{\alpha-y}x^{m-1}e^{-\mu x}\,dx$

Question

Let $f(x)$ be the function defined by $$f(x)=x^{m-1}e^{-\mu x}$$

where $m$ is positive integer and $\mu$ is positive real number.

I would like to compute

$$ I=\int_{x=0}^{\alpha-y}f(x)dx=\int_{x=0}^{\alpha-y}x^{m-1}e^{-\mu x}dx. $$

where $y\in[0,\infty[$

So is it $$I=\int_{x=0}^{\alpha-y}f(x)=\frac{\gamma\Big(m,\mu(\alpha-y)\Big)}{\mu^m}$$

or do we need the condition that $y\leq \alpha$?

Thanks.

Answer 1

$$I=\int_0^{\alpha-y}x^{m-1}e^{-\mu x}dx$$ by letting $u=\mu x$ we get: $$I=\int_0^{\mu(\alpha-y)}\left(\frac{u}{\mu}\right)^{m-1}e^{-u}\frac{du}{\mu}=\mu^{-m}\int_0^{\mu(\alpha-y)}u^{m-1}e^{-u}du$$ and since the lower incomplete gamma function is defined as: $$\gamma(s,x)=\int_0^xt^{s-1}e^{-t}dt$$ we can say that: $$I=\mu^{-m}\gamma\bigl(m,\mu(\alpha-y)\bigr)$$ Which agrees with what you have said. I believe if we were to extend it from $x$ to $z\in\mathbb{C}$ there are then conditions added, where $\Re(m)>1$, however this is similar as is for the real domain. As for the limits of the integral, the only notatable point I think is that as $(\alpha-y)\to-\infty,I\to\infty$ and it will give negative values in $x$, giving a complex result, however this depends on the value of $m$

Answer 2

For negative $\alpha -y$, the integral is taken over negative values of $x$.

For negative $x$ , the integrand function is defined over the reals, iff $m$ is a integer, as you specify.

So you do not have any limitation onto $y$, when this is real.