Sunday, 21 December 2014

calculus - How to prove the closed form of the integral intfracdxprodnr=0(x+r)




I want to derive a closed formula for the integral In=dxnr=0(x+r)



On writing out first few terms we get



For n=0, I0=ln|x|+C
For n=1, I1=ln|x|ln|x+1|+C
For n=2 I_2=\frac {1}{2!}\left(\sum_{r=0}^2 (-1)^r\binom {2}{r} \ln \vert x+r\vert\right)+C
For n=3 I_3=\frac {1}{3!}\left(\sum_{r=0}^3 (-1)^r\binom {3}{r} \ln \vert x+r\vert\right)+C




Hence for generalized n we have I_n=\frac {1}{n!}\left(\sum_{r=0}^n (-1)^r\binom {n}{r} (\ln \vert x+r\vert)\right)+C



Now this is just an observation but I want to prove that it is correct. I have tried lot of methods but not useful. Partial fractions would have been most useful but would go out on tedious task which is nearly impossible. Also integration by parts won't help nor any trig substitution. So any ideas are welcome. And ya, this is not a homework question, it's a question which I just saw in a integral challenge paper.


Answer



\newcommand{\bbx}[1]{\,\bbox[15px,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} \int{\dd x \over \prod_{r = 0}^{n}\pars{x + r}} & = \int{\dd x \over \Gamma\pars{x + n + 1}/\Gamma\pars{x}} = {1 \over n!} \int{\Gamma\pars{x}\Gamma\pars{n + 1}\over \Gamma\pars{x + n + 1}}\,\dd x \\[5mm] & = {1 \over n!}\int\int_{0}^{1}t^{x - 1}\pars{1 - t}^{n}\,\dd t\,\dd x = {1 \over n!}\int_{0}^{1}\pars{1 - t}^{n}\int t^{x - 1}\,\dd x \\[5mm] & = {1 \over n!}\int_{0}^{1}\pars{1 - t}^{n} \bracks{{t^{x - 1} \over \ln\pars{t}} + \,\mrm{A}\pars{t}}\,\dd x \\[5mm] & = {1 \over n!}\int_{0}^{1}{t^{x - 1}\pars{1 - t}^{n} \over \ln\pars{t}}\,\dd t + {1 \over n!}\int_{0}^{1}{t^{x - 1}\,\mrm{A}\pars{t} \over \ln\pars{t}}\,\dd t \end{align}




\ds{\mrm{A}\pars{t}} is an integration "constant" ( it doesn't depend on \ds{x} but, in general, it depends on \ds{t} ).



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