Integral equation proof by induction

Question

I am trying to solve the following problem:

We define the following integral for $n \in \mathbb{N}$ and $x \in \mathbb{R}_+^*$:

$$ \mathcal{I}(n,x) = \mathcal{I}_x(n) = \int_0^Tt^ne^{-xt}dt $$

Prove that:

$$ \begin{align} \mathcal{I}_x(n) & = \frac{n!}{x^{n+1}} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}} \\[12pt] & = \frac{n!}{x^{n+1}} \left(1 - e^{-xT}\sum_{k=0}^n{\frac{(xT)^k}{k!}}\right)\end{align} $$

My try through induction:

• Case $n=0$: $$ \begin{align} \mathcal{I}_x(0) & = \int_0^Te^{-xt}dt \\[12pt] & = \left[-\frac{e^{-xt}}{x} \right]_{t=0}^{t=T} \\[12pt] & = \frac{1}{x} - e^{-xT}\frac{1}{x} \end{align} $$ Hence the statement is true for $n=0$.

• Assume case $n$ is true. Integration by parts gives us: $$ \begin{align} \mathcal{I}_x(n+1) & = \int_0^Tt^{n+1}e^{-xt}dt \\[12pt] & = \left[-\frac{t^{n+1}e^{-xt}}{x} \right]_{t=0}^{t=T} + \frac{n+1}{x}\int_0^T{t^ne^{-xt}dt} \\[12pt] & = -\frac{T^{n+1}e^{-xT}}{x} + \frac{n+1}{x}\mathcal{I}_x(n) \\[12pt] & = -\frac{T^{n+1}e^{-xT}}{x} + \frac{n+1}{x}\left( \frac{n!}{x^{n+1}} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}} \right) \\[12pt] & = \frac{(n+1)!}{x^{(n+1)+1}}-\frac{T^{n+1}e^{-xT}}{x} - \frac{n+1}{x}e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}} \\[12pt] & = \frac{(n+1)!}{x^{(n+1)+1}}-\frac{T^{n+1}e^{-xT}}{x} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{(n+1)-k+1}\frac{(n+1)!}{k!}} \\[12pt] & = \frac{(n+1)!}{x^{(n+1)+1}}- e^{-xT}\sum_{k=0}^{n+1}{T^k\left(\frac{1}{x}\right)^{(n+1)-k+1}\frac{(n+1)!}{k!}} \end{align} $$

  Hence if the statement holds for $n$ it also holds for $n+1$, which completes the proof.

I was expecting a simpler expression rather than the complicated sum. Does my reasoning seem right here? Is there any mistake? Is there a way to simplify further the final expression?

Answer 1

After looking around, it seems my result is accurate: for $n \in \mathbb{N}^*$ and $T \in \mathbb{R}_+^*$, let $\gamma(n,T)$ and $\Gamma(n,T)$ be the lower and upper incomplete gamma functions respectively:

$$ \begin{align} & \gamma(n,T) = \int_0^Tu^{n-1}e^{-u}du \\[12pt] & \Gamma(n,T) = \int_T^{\infty}u^{n-1}e^{-u}du \end{align} $$

Wikipedia's page on incomplete gamma functions states that $-$ if $n \in \mathbb{N}^*$:

$$ \Gamma(n,T) = (n-1)! \ e^{-T}\sum_{k=0}^{n-1} \frac{T^k}{k!} $$

Given $\Gamma(n) = \gamma(n,T) + \Gamma(n,T)$ and $\Gamma(n) = (n-1)!$ $-$ provided $n$ is a positive integer $-$ it comes:

$$ \gamma(n,T) = \Gamma(n) - \Gamma(n,T) = (n-1)! \left(1- e^{-T}\sum_{k=0}^{n-1} \frac{T^k}{k!}\right)$$

Hence, making the dependence on $T$ explicit by noting my integral $\mathcal{I}$ as $\mathcal{I}(n,x,T)$:

$$ \gamma(n,T) = \mathcal{I}(n-1,1,T) $$

A simple change of variable $-$ $u = xt$ $-$ in $\mathcal{I}(n,x,T)$ yields:

$$ \mathcal{I}(n,x,T) = \frac{1}{x^{n+1}}\int_0^{xT} u^n e^{-u}du = \frac{\gamma(n+1,xT)}{x^{n+1}}$$

Plugging this back into the sum representation of $\gamma(\cdot,\cdot)$:

$$ \mathcal{I}(n,x,T) = \frac{\gamma(n+1,xT)}{x^{n+1}} = \frac{n!}{x^{n+1}} \left(1- e^{-xT}\sum_{k=0}^n \frac{(xT)^k}{k!}\right)$$

which corresponds to my initial representation of $\mathcal{I}_x(n)$ so I guess it must be right $-$ given I now have 2 proofs.