Integral from zero to infinity of $\int_0^{\infty}\frac{(1-e^{-\lambda z})}{\lambda^{a+1}} d \lambda$

Question

I know that the value of the integral is as follows $$\int_0^{\infty}\frac{(1-e^{-\lambda z})}{\lambda^{a+1}} d \lambda =z^a \frac{\Gamma(1-a)}{a}$$

However, how exactly the integral is calculated? How one proves that the integral even exists? Why does the integral only exist for $0<a<1$?

Answer 1

Existence: at $\lambda=\infty$, $a>0$ guarantees that the integrand is below $1/\lambda{1+a}$ which is integrable. At $\lambda=0$, $$ \int_0^\delta\frac{(1-e^{-\lambda z})}{\lambda^{a+1}}\,d\lambda=\int_0^\delta\frac{\lambda z+O(\lambda^2)}{\lambda^{a+1}}\,d\lambda=\int_0^\delta\frac{z}{\lambda^a}+O(\lambda^{1-a})\,d\lambda. $$ The first term requirest $a<1$ for convergence, and the second one $a<2$. So $0<a<1$ are precisely the values of $a$ for which your integral exists.

Regarding the calculation, I wouldn't know how to find the formula, but here is a way to justify it. It suffices to use differentiation under the integral sign: we have $$ \frac d{dz}\,\int_0^\infty\frac{(1-e^{-\lambda z})}{\lambda^{a+1}}\,d\lambda=\int_0^\infty\frac{\lambda\,e^{-\lambda z}}{\lambda^{a+1}}\,d\lambda=\int_0^\infty\lambda^{-a}\,e^{-\lambda z}\,d\lambda=\int_0^\infty z^{a-1}\,t^{-a}\,e^{-t}\,dt=z^{a-1}\Gamma(1-a)=\frac d{dz}\,\left(\frac{z^a}a\,\Gamma(1-a)\right). $$ So the two sides of your equality differ by a constant $c$. When $z=0$ both sides equal to $0$, so $c=0$.

Answer 2

$$ \int_{0}^{\infty} \frac{ 1-e^{-z \lambda}}{\lambda^{a+1}} \ d \lambda = \frac{1}{\Gamma(a+1)} \int_{0}^{\infty} \int_{0}^{\infty} (1-e^{-z \lambda}) t^{a} e^{- \lambda t} \ \ dt \ d \lambda$$

$$= \frac{1}{\Gamma(a+1)} \int_{0}^{\infty} \int_{0}^{\infty} t^{a} \Big( e^{- \lambda t} - e^{-(z+t)\lambda} \Big) \ d \lambda \ d t$$

$$ = \frac{1}{\Gamma(a+1)} \int_{0}^{\infty} t^{a} \Big( \frac{1}{t} - \frac{1}{z+t} \Big) \ dt= \frac{z}{\Gamma(a+1)} \int^{\infty}_{0} \frac{t^{a-1}}{z+t} \ dt$$

$$ = \frac{z^{a}}{\Gamma(a+1)} \int_{0}^{\infty} \frac{u^{a-1}}{1+u} \ du = \frac{z^{a}}{\Gamma(a+1)} B(a,1-a) $$

$$= \frac{z^{a}}{\Gamma(a+1)} \Gamma(a) \Gamma(1-a) = z^{a} \frac{\Gamma(1-a)}{a}$$

Answer 3

I can tell you why the integral only exists for $a \in (0.1)$. For $\lambda$ near $0$, the integrand approaches$\lambda z/\lambda^{a+1} = z \lambda^{-a}$, the primitive of which is $z/(1-a) \lambda^{1-a}$, which is finite at $\lambda=0$ only when $a \lt 1$. (When $a=1$, the above primitive is not valid and it is rather $\log{\lambda}$, which is not finite $\lambda=0$.) For large $\lambda$ on the other hand, the primitive behaves as $\lambda^{-a}/a$, which is only finite in the limit as $\lambda \to \infty$ when $a \gt 0$. The intersection of these two conditions is $a \in (0,1)$.

Answer 4

Making use of: $$\int_0^z e^{-\lambda t} dt = \frac{1 - e^{-\lambda t}}{\lambda}$$ and changing order of integration we can obtain: $$\int_0^z dt \int_0^{+\infty} \lambda^{-a}e^{-\lambda t} d\lambda \stackrel{x = \lambda t}{=} \int_0^z t^{a-1} \,dt\int_0^{+\infty} x^{-a} e^{-x} dx = \Gamma (1-a) \int_0^z t^{a-1} \,dt = \frac{z^a \Gamma (1-a)}{a}$$