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Let $s\in(0,1)$. I managed to prove that \begin{align} \frac{1}{\lambda^s}=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}e^{-\lambda t}dt,\qquad\lambda>0. \end{align} It directly follows from the definition of the Gamma function as \begin{align} \Gamma(x)=\int_0^\infty x^{s-1}e^{-x} dx, x>0 \end{align} and a change of variables. Now I try to prove that \begin{align} \lambda^s=\frac{1}{|\Gamma(-s)|}\int_0^\infty t^{-(s+1)}(1-e^{-\lambda t})dt,\qquad\lambda>0. \end{align} But I'm stuck.

Attempt: We can define the gamma function for $-s$ as \begin{align} \Gamma(-s):=\frac{-1}{s}\Gamma(1-s):=-\frac{\lambda^{1-s}}{s}\int_0^\infty t^{-s}e^{-\lambda t} \end{align} I tried Integration-by-parts, \begin{align} \int u dv=uv-\int v du. \end{align} I chose $u=t^{-s}$ and $dv=-\lambda e^{-\lambda t}$. But $uv=t^{-s}e^{-\lambda t}$ which diverges as $t$ goes to zero.

DonQuixote
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2 Answers2

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Fix some $s\in(0,1)$, and consider the function defined for $\lambda>0$ given by $$ f(\lambda) := \lambda^s + \frac 1{\Gamma(-s)}\int_0^\infty t^{-s}(1-e^{-\lambda t})\;\frac{dt}t\ . $$ Then $f(0)=0$, and $$ \begin{aligned} f'(\lambda) &= s\lambda^{s-1} + \frac 1{\Gamma(-s)}\int_0^\infty t^{1-s}e^{-\lambda t}\;\frac{dt}t \\ &= s\left( \lambda^{s-1} - \frac 1{-s\Gamma(-s)}\int_0^\infty t^{1-s}e^{-\lambda t}\;\frac{dt}t \right) \\ &= s\left( \frac 1{\lambda^{\color{blue}{1-s}}} - \frac 1{\Gamma(\color{blue}{1-s})} \int_0^\infty t^{\color{blue}{1-s}}e^{-\lambda t}\;\frac{dt}t \right) \\ &=0 \ , \end{aligned} $$ by using the first relation from the posted question (for $\color{blue}{1-s}\in(0,1)$ instead of $s$). So $f$ is identically zero.

$\square$

dan_fulea
  • 32,856
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You have $$|\Gamma(-s)|=\frac{1}{s}\Gamma(1-s)=\frac{\lambda^{1-s}}{s}\int_0^\infty t^{-s}e^{-\lambda t}dt.$$ Thus, you have $$|\Gamma(-s)|\lambda^s=\frac{\lambda}{s}\int_0^\infty t^{-s}e^{-\lambda t}.dt$$ Therefore, what you try to prove is that $$\int_0^\infty t^{-(s+1)}(1-e^{-\lambda t})dt=\frac{\lambda}{s}\int_0^\infty y^{-s}e^{-\lambda t}.dt$$ By doing by parts (you don't have your problem at $0$ anymore since you have the $1$ now): $f(t)=1-e^{-\lambda t}$ -> $f'(t) = \lambda e^{-\lambda t}$ and $g'(t)=t^{-(s+1)}$ -> $g(t) = \frac{t^{-s}}{-s}$, you find easily the previous formula.

SacAndSac
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