# Closed form for $\int_0^t(x+c)^p(1-2x)^{N-1}\text{d}x$

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. $$$$\begin{split} I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}-c^{p+1}\right]+\\[5pt] &\quad+\frac{2(N-1)}{p+1}\int_0^t(x+c)^{p+1}(1-2x)^{N-2}\text{d}x=\dots\end{split}$$$$ By iterating $$N$$ times the above step the starting integral can be rewritten as a sum $$$$\tag{1}\label{eq1} I=\Gamma(N)\Gamma(p+1)\sum_{j=1}^N\frac{2^{j-1}}{\Gamma(N-j+1)\Gamma(p+j+1)}\left[(t+c)^{p+j}(1-2t)^{N-j}-c^{p+j}\right].$$$$ 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 $$$$\begin{split} I&=\frac{1}{p+1}\left[(t+c)^{p+1}(1-2t)^{N-1}{}_2F_1\left(1,1-N;p+2;-\frac{2(t+c)}{1-2t}\right)\right.\\[5pt] &\left.\quad-c^{p+1}{}_2F_1(1,1-N;p+2;-2c)\right].\end{split}$$$$

• Is this homework ? Commented Jun 2, 2019 at 4:27
• 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. Commented Jun 2, 2019 at 8:58
• 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. Commented Jun 2, 2019 at 9:07
• 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. Commented Jun 2, 2019 at 21:16
• 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) Commented Jun 2, 2019 at 21:28

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 $$\sf 0. $$\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.

• 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! Commented Jun 2, 2019 at 22:23
• 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 $p+1\leftrightarrow N$). Commented Jun 3, 2019 at 9:52
• @ARWarrior Definetly! Thanks for noticing that. Commented Jun 3, 2019 at 10:07
• (+1), but why use serif and typewriter fonts? Commented Nov 7, 2021 at 12:59
• @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). Commented Nov 7, 2021 at 13:14

In fact you don't need the hypergeometric function for the cases of natural number $$N$$.

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

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

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

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

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

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

• 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). Commented Nov 4, 2019 at 23:09
• 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). Commented Nov 4, 2019 at 23:19