I need help in solving following integral $$\int_u^{\infty}x\left(1-x^{km}(a+x^k)^{-m}\right)dx$$ where $a,m,u$ are positive real values and $k>2$. I have seen in one of the paper that the result can be written in the form of Gauss Hypergeometric function. I will be very thankful to you for your help.

This result has been obtained using a CAS.

Concerning the antiderivative $$I=\int x\left(1-x^{km}(a+x^k)^{-m}\right)dx$$ $$I=\frac{x^2}{2}-\frac{x^{k m+2} \left(a+x^k\right)^{1-m}}{a (k m+2)}\,\, _2F_1\left(1,\frac{k+2}{k};\frac{2}{k}+m+1;-\frac{x^k}{a}\right)$$ Concerning the limit when $x\to\infty$, I give below a few values $$\left( \begin{array}{ccc} a & m & k & \text{limit} \\ 1 & 1 & 3 & \frac{2 \pi }{3 \sqrt{3}} \\ 1 & 2 & 3 & \frac{10 \pi }{9 \sqrt{3}} \\ 1 & 3 & 3 & \frac{40 \pi }{27 \sqrt{3}} \\ 1 & 4 & 3 & \frac{440 \pi }{243 \sqrt{3}} \\ 1 & 5 & 3 &\frac{1540 \pi }{729 \sqrt{3}} \\ 2 & 1 & 3 &\frac{2\ 2^{2/3} \pi }{3 \sqrt{3}} \\ 2 & 2 & 3 &\frac{10\ 2^{2/3} \pi }{9 \sqrt{3}} \\ 2 & 3 & 3 &\frac{40\ 2^{2/3} \pi }{27 \sqrt{3}} \\ 2 & 4 & 3 &\frac{440\ 2^{2/3} \pi }{243 \sqrt{3}} \\ 2 & 5 & 3 &\frac{1540\ 2^{2/3} \pi }{729 \sqrt{3}} \\ 1 & 1 & 4 & \frac{\pi }{4} \\ 1 & 2 & 4 & \frac{3 \pi }{8} \\ 1 & 3 & 4 & \frac{15 \pi }{32} \\ 1 & 4 & 4 & \frac{35 \pi }{64} \\ 1 & 5 & 4 & \frac{315 \pi }{512} \\ 2 & 1 & 4 & \frac{\pi }{2 \sqrt{2}} \\ 2 & 2 & 4 & \frac{3 \pi }{4 \sqrt{2}} \\ 2 & 3 & 4 & \frac{15 \pi }{16 \sqrt{2}} \\ 2 & 4 & 4 & \frac{35 \pi }{32 \sqrt{2}} \\ 2 & 5 & 4 & \frac{315 \pi }{256 \sqrt{2}} \end{array} \right)$$ For $k=2$, the limit is effectively $+\infty$.

  • Thank you for your answer. Actually this question was a result of math.stackexchange.com/questions/1954619/… . I have added my attempt in that question. Can you please tell me where I am doing that question wrong? Thanks in advance – Frank Moses Oct 12 '16 at 5:40
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    @FrankMoses. I remember the previous post. – Claude Leibovici Oct 12 '16 at 5:46

Hint: \begin{equation} \int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x \end{equation}

Let $y = -\frac{x^{k}}{a}$ \begin{align} \int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x &= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \int (1-y)^{-m} y^{m-1+2/k} \mathrm{d}y \\ &= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \mathrm{B}_{y}\left(m+\frac{2}{k},1-m\right) \\ &= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \frac{y^{m+2/k}}{m+2/k} {}_{2}\mathrm{F}_{1}\left(m+\frac{2}{k},m;m+1+\frac{2}{k};y\right) \\ &= \frac{1}{km+2} \frac{1}{a^{m}} x^{km+2} {}_{2}\mathrm{F}_{1}\left(m+\frac{2}{k},m;m+1+\frac{2}{k};-\frac{x^{k}}{a}\right) \end{align}

Notes: 1. \begin{equation} \mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t \end{equation} is the incomplete beta function.

2. \begin{equation} \mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} {}_{2}\mathrm{F}_{1}(p,1-q;p+1;z) \end{equation} is the incomplete beta function in terms of Gauss's hypergeometric function.

  • Thank you for your answer for the second part in the paranthesis. I do not understand how to solve the $\int_u^{\infty}xdx$ part. Actually this question was a result of math.stackexchange.com/questions/1954619/… . I have added my attempt in that question. Can you please tell me where I am doing that question wrong? Thanks in advance – Frank Moses Oct 12 '16 at 5:39

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