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I am trying to find a closed form for the integral $$I\equiv\int_0^t(x+c)^p(1-2x)^{N-1}\text{d}x,$$ where $N\in\mathbb{N}$, $p>0$, $c\ge 0$ and $t\in\left(0,\frac{1}{2}\right)$.

I thought to evaluate the integral proceeding by parts, in order to lower the integer power of the second factor in the integrand. \begin{equation}\begin{split} I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}-c^{p+1}\right]+\\[5pt] &\quad+\frac{2(N-1)}{p+1}\int_0^t(x+c)^{p+1}(1-2x)^{N-2}\text{d}x=\dots\end{split} \end{equation} By iterating $N$ times the above step the starting integral can be rewritten as a sum \begin{equation}\tag{1}\label{eq1} I=\Gamma(N)\Gamma(p+1)\sum_{j=1}^N\frac{2^{j-1}}{\Gamma(N-j+1)\Gamma(p+j+1)}\left[(t+c)^{p+j}(1-2t)^{N-j}-c^{p+j}\right]. \end{equation} I wonder whether this is the best result one can hope for, or if further simplifications can be performed, maybe by evaluating the sum in a closed form or by proceeding in a different way from the beginning in the solution of the integral.


Edit: after the mention of hypergeometric functions in the comments, I managed to rewrite the above expression in their terms. To show this I will consider only the second term in the square brakets, but the procedure is the same for the first one. $$S_1\equiv\sum_{j=1}^N\frac{(2c)^j}{\Gamma(N-j+1)\Gamma(p+j+1)}=\left(\sum_{j=1}^{\infty}-\sum_{j=N+1}^{\infty}\right)\frac{(2c)^j}{\Gamma(N-j+1)\Gamma(p+j+1)}$$ With the substitutions $j\rightarrow j-1$ and $j\rightarrow j-(N+1)$ in the first and second sum respectively I got $$S_1=\sum_{j=0}^{\infty}\left[\frac{(2c)^{j+1}}{\Gamma(N-j)\Gamma(p+j+2)}-\frac{(2c)^{j+N+1}}{\Gamma(-j)\Gamma(p+j+N+2)}\right],$$ where the second fraction vanishes due to the fact that $1/\Gamma(-j)=0\;\;\forall j\in\mathbb{N}$. Using now the relation $$\Gamma(\epsilon-n)=(-1)^{n-1}\frac{\Gamma(-\epsilon)\Gamma(1+\epsilon)}{\Gamma(n+1-\epsilon)}$$ with the identifications $\epsilon=N$ and $n=j$, I found $$S_1=-\frac{2c}{\Gamma(-N)\Gamma(N+1)}\sum_{j=0}^{\infty}\frac{\Gamma(j+1-N)}{\Gamma(p+j+2)}(-2c)^j.$$ Thanks to the formula $\Gamma(z+1)=z\Gamma(z)$ I wrote $\Gamma(-N)\Gamma(N+1)=\Gamma(1-N)\Gamma(N)$, and it is quite easy using the definition of the hypergeometric function ${}_2F_1(a,b;c;z)$ to verify that $$S_1=\frac{2c}{\Gamma(N)\Gamma(p+2)}{}_2F_1(1,1-N;p+2;-2c).$$ Exactly in the same way I found that $$\sum_{j=1}^N\frac{2^j}{\Gamma(N-j+1)\Gamma(p+j+1)}\left(\frac{t+c}{1-2t}\right)^j=\frac{2}{\Gamma(N)\Gamma(p+2)}\frac{t+c}{1-2t}{}_2F_1\left(1,1-N;p+2;-\frac{2(t+c)}{1-2t}\right).$$ Putting together these results the integral of interest becomes \begin{equation}\begin{split} I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}{}_2F_1\left(1,1-N;p+2;-\frac{2(t+c)}{1-2t}\right)\right.\\[5pt] &\left.\quad-c^{p+1}{}_2F_1(1,1-N;p+2;-2c)\right].\end{split} \end{equation}

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  • $\begingroup$ Is this homework ? $\endgroup$ Commented Jun 2, 2019 at 4:27
  • $\begingroup$ No it isn't, this is part of a broader calculation I am interested in. The subsequent step should be summing over $k=1,\dots,\lfloor(N-1)/2\rfloor$, this is why I was looking for a way to simplify my result or to approach the integral in a different way. $\endgroup$
    – ARWarrior
    Commented Jun 2, 2019 at 8:58
  • $\begingroup$ I bet that the result involves somewhere beta functions. The antiderivative is more than likely an hypergeometric function. The problem is probably later. If you give more context, I could try using a few CAS to see what is coming out. $\endgroup$ Commented Jun 2, 2019 at 9:07
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    $\begingroup$ I guess you can do something like this. Denote$$I=2^{N-1}\int_0^t (x+c)^p\left(\frac12-x\right)^{N-1}dx$$ $$\text{let } \frac{\frac12-x}{\frac12+c}=y\Rightarrow x=\frac12-\frac{y}{2}-cy $$ $$I=2^{N-1}\left(c+\frac12\right)\int_{\frac{\frac12-t}{\frac12+c}}^{\frac{\frac12}{\frac12+c}} \left((1-y)\left(c+\frac12\right)\right)^p\left(y\left(c+\frac12\right)\right)^{N-1}dy$$ $$=2^{N-1}\left(c+\frac12\right)^{p+N} \int_{\frac{\frac12-t}{\frac12+c}}^{\frac{\frac12}{\frac12+c}} (1-y)^p y^{N-1}dy$$ Now split the interval into two parts so that you can apply incomplete beta function. $\endgroup$
    – Zacky
    Commented Jun 2, 2019 at 21:16
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    $\begingroup$ I just saw your comment because I was editing my post. I will try to finish the calculation from myself. (Whether it is good or bad maybe the result will be in a form more suitable for the subsequent steps of my calculations, so thank you very much) $\endgroup$
    – ARWarrior
    Commented Jun 2, 2019 at 21:28

2 Answers 2

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So just to have everything here, let me show a general approach for the following $$\sf I=\int_e^f (a+x)^p(b-x)^qdx$$ Either one of the substitution $\sf \frac{b-x}{b+a}=y$ or $\sf \frac{a+x}{b+a}=y$ should work fine as a start.

I'll take the first one and get $\sf x=b-y(a+b)\Rightarrow dx=-(a+b)dy$.

Also for better view, denote $\sf \frac{b-f}{b+a}=k$ and $\sf \frac{b-e}{b+a}=n$ and assume $\sf 0<e<f$. $$\sf \Rightarrow I=(a+b)\int_{k}^{n}(a+b-y(a+b))^p(b-b+y(a+b))^qdy$$ $$\sf =(a+b)^{p+q+1}\int_k^n(1-y)^py^qdy=(a+b)^{p+q+1} \left(\int_0^n-\int_0^k\right)$$ $$\sf =(a+b)^{p+q+1}\left(B\left(\frac{b-e}{b+a};q+1,p+1\right)-B\left(\frac{b-f}{b+a};q+1,p+1\right)\right)$$ Where $\sf B(z;\alpha,\beta)$ is the Incomplete Beta function also called Chebyshev Integral.


In particular your integral is equal to: $$\tt 2^{N-1}\int_0^t (x+c)^p(1/2-x)^{N-1}dx$$ $$\tt =2^{N-1}\left(c+\frac12\right)^{p+N}\left(B\left(\frac{1}{1+2c};N,p+1\right)-B\left(\frac{1-2t}{1+2c};N,p+1\right)\right)$$ Hopefully I haven't done any computation mistakes, but feel free to ask if anything is unclear.

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    $\begingroup$ I completed it too just now. I think this is the simplest form one can hope for starting from the given integral. Again thank you! $\endgroup$
    – ARWarrior
    Commented Jun 2, 2019 at 22:23
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    $\begingroup$ Just a little note. According to the definition of the incomplete beta function I think that the parameters $\alpha$ and $\beta$ should be inverted in the solution both of the general and the specific case (i.e. $p+1\leftrightarrow q+1$ and $p+1\leftrightarrow N$). $\endgroup$
    – ARWarrior
    Commented Jun 3, 2019 at 9:52
  • $\begingroup$ @ARWarrior Definetly! Thanks for noticing that. $\endgroup$
    – Zacky
    Commented Jun 3, 2019 at 10:07
  • $\begingroup$ (+1), but why use serif and typewriter fonts? $\endgroup$ Commented Nov 7, 2021 at 12:59
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    $\begingroup$ @TymaGaidash I guess we all have those periods in life where we use random stuff like too much emojis, chat language and so on (because we think those look cool). $\endgroup$
    – Zacky
    Commented Nov 7, 2021 at 13:14
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In fact you don't need the hypergeometric function for the cases of natural number $N$.

$\int_0^t(x+c)^p(1-2x)^{N-1}~dx$

$=\int_c^{t+c}x^p(2c+1-2x)^{N-1}~dx$

$=\int_c^{t+c}x^p\sum\limits_{m=0}^{N-1}C_m^{N-1}(2c+1)^{N-m-1}(-1)^m2^mx^m~dx$

$=-\int_c^{t+c}\sum\limits_{m=1}^NC_{m-1}^{N-1}(-1)^m2^{m-1}(2c+1)^{N-m}x^{m+p-1}~dx$

$=-\left[\sum\limits_{m=1}^N\dfrac{(-1)^m2^{m-1}(2c+1)^{N-m}(N-1)!x^{m+p}}{(m-1)!(N-m)!(m+p)}\right]_c^{t+c}$

$=\sum\limits_{m=1}^N\dfrac{(-1)^m2^{m-1}(2c+1)^{N-m}(N-1)!(c^{m+p}-(t+c)^{m+p})}{(m-1)!(N-m)!(m+p)}$

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  • $\begingroup$ I've edited your answer to correct a small algebraic mistake. Apart from that, your answer provides a more elegant calculation than a series of integration by parts to obtain my expression for $I$ given above in Eq.$(1)$. Moreover, by comparison it implies that $$\sum_{k=1}^N\frac{2^{k-1}}{\Gamma(N-k+1)\Gamma(p+k+1)}[(t+c)^{p+k}(1-2t)^{N-k}-c^{p+k}]=\sum_{k=1}^N\frac{(-2)^{k-1}}{\Gamma(N-k+1)\Gamma(k)\Gamma(p+1)(k+p)}[(t+c)^{p+k}-c^{p+k}](1+2c)^{N-k}$$ (maybe this is straightforward but I'm struggling to show why). $\endgroup$
    – ARWarrior
    Commented Nov 4, 2019 at 23:09
  • $\begingroup$ I'll try to figure out if the new expression that you provided here can be further simplified (the hypergeometric representation which I mentioned was just an attempt in this direction, although probably a dead end). $\endgroup$
    – ARWarrior
    Commented Nov 4, 2019 at 23:19

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