Need help calculating a definite integral

Question

I'm trying to calculate the following integral:

$$ \int_0^1 e^{-\lambda(1-x)} (1-x)^{n-1} x^{k-n} dx $$ It seems like kind of a combination of Gamma and Beta function. I'm suspecting that it has something to do with both of them, however I can't see the connection.

Any hint would be welcome!

Answer 1

The integrand is the product of an exponential and a polynomial. You can pull out a factor $e^{-\lambda}$ and compute the antiderivative explicitly.

Let it be $Q(x)e^{\lambda x}$, such that

$$(Q(x)e^{\lambda x})'=(\lambda Q(x)+Q'(x))e^{\lambda x}=P(x)e^{\lambda x}=\sum_{j=0}^{n-1}\binom{n-1}j(-1)^jx^{j+k-n}e^{\lambda x}.$$

The unknown coefficients of $Q$ form a bidiagonal system of equations, which is straigthforward to solve ($\lambda q_i+iq_{i-1}=p_i$), and the definite integral is

$$Q(1)-Q(0)e^{-\lambda},$$ where $Q(1)$ is the sum of the coefficients of $Q$.

Answer 2

Note that $$I\left(\lambda,k,n\right)=\int_{0}^{1}e^{-\lambda\left(1-x\right)}\left(1-x\right)^{n-1}x^{k-n}dx$$ $$=\int_{0}^{1}e^{-\lambda x}x^{n-1}\left(1-x\right)^{k-n}dx=\sum_{m\geq0}\frac{\left(-1\right)^{m}\lambda^{m}}{m!}\int_{0}^{1}x^{m+n-1}\left(1-x\right)^{k-n}dx$$ $$=\sum_{m\geq0}\frac{\left(-1\right)^{m}\lambda^{m}}{m!}B\left(m+n,k-n+1\right)$$ and since $$B\left(x+1,y\right)=B\left(x,y\right)\frac{x}{x+y}\tag{1}$$ we have, iterating $(1)$ $m$ times, that $$I\left(\lambda,k,n\right)=B\left(n,k-n+1\right)\sum_{m\geq0}\frac{\left(-1\right)^{m}\lambda^{m}}{m!}\frac{\left(n\right)_{m}}{\left(k+1\right)_{m}}=\color{red}{B\left(n,k-n+1\right) \,_{1}F_{1}\left(n;k+1;-\lambda\right)}$$ where $_{1}F_{1}\left(a;b;z\right)$ is the Kummer confluent hypergeometric function. So Claude Leibovici's suspicions were correct.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#efe,15px]{\ds{\int_{0}^{1}\expo{-\lambda\pars{1 - x}} \pars{1 - x}^{n - 1}x^{k - n}\,\dd x}} = \expo{-\lambda}\int_{-1/2}^{1/2}\expo{\lambda\pars{x + 1/2}} \pars{{1 \over 2} - x}^{n - 1}\pars{x + {1 \over 2}}^{k - n}\,\dd x \\[5mm] = &\ \expo{-\lambda/2}\int_{-1/2}^{1/2}\expo{\lambda x}\, {1 \over 2^{n - 1}}\pars{1 - 2x}^{n - 1}\,{1 \over 2^{k - n}}\, \pars{2x + 1}^{k - n}\,{1 \over 2}\,2\,\dd x \\[5mm] = &\ 2^{-k}\expo{-\lambda/2}\int_{-1}^{1}\expo{\lambda x/2}\, \pars{1 + x}^{k - n}\pars{1 - x}^{n - 1}\,\dd x \\[5mm] = &\ 2^{-k}\expo{-\lambda/2}\, {\Gamma\pars{k - n}\Gamma\pars{n} \over \Gamma\pars{k}\lambda^{k/2}2^{1 - k}}\, \,\mrm{M}_{n - k/2,k/2 - 1/2}\pars{\lambda} \\[5mm] = & \bbx{\ds{{1 \over 2}\,\lambda^{-k/2}\expo{-\lambda/2}\, {\Gamma\pars{k - n}\Gamma\pars{n} \over \Gamma\pars{k}}\, \,\mrm{M}_{n - k/2,k/2 - 1/2}\pars{\lambda}}}\,,\qquad \Re\pars{k \over 2} > \verts{\Re\pars{n - {k \over 2}}} \end{align}

$\ds{\mrm{M}_{\kappa,\nu}}$ is a Whittaker Function. Note that $\ds{{\Gamma\pars{k - n}\Gamma\pars{n} \over \Gamma\pars{k}} = \,\mrm{B}\pars{k - n,n}}$ where $\ds{\,\mrm{B}}$ is the Beta Function.