# Improper integral and hypergeometric function

I want to know how does the following equation is true (I have checked for different values in Wolfram Alpha and the answer obtained from both sides are equal) $$\int_u^\infty \left[1-\frac{1}{\left(1+\frac{a}{x^k}\right)^m}\right] x \, dx = -\frac{u^2}{2} + \frac{u^2}{2} \left[_2F_1(m;-\frac{2}{k};1-\frac{2}{k};\frac{a}{u^k})\right]$$where $a,m,u$ are all positive values and $k>2$.

My attempt:

In my attempt I can use $v=\left(\frac{u}{x}\right)^k$ I can find $$\text{when } x=u \to v=1 \text{ and when } x=\infty \to v=0$$ $$\frac{dx}{dz}=-\frac{u}{k}v^{-\frac{1}{k}-1}$$ therefore with substitution we can have $$\int_u^\infty\left[1-\frac{1}{\left(1+\frac{a}{x^k} \right)^m} \right] x \, dx=\int_0^1\left[1-\frac{1}{\left(1+\frac{av}{u^k}\right)^m}\right]uv^{-\frac{1}{k}}\frac{u}{k}v^{-\frac{1}{k}-1}dv$$ $$\int_u^\infty\left[1-\frac{1}{\left(1+\frac{a}{x^k} \right)^m} \right] x \, dx = \int_0^1 \left[1-\left(\frac{1}{1+\frac{av}{u^k}}\right)^m\right](\frac{u^2}{k}) v^{-\frac{2}{k}-1} \,dv$$ now if somehow there is $v^{\frac{2}{k}-1}$ instead of $v^{-\frac{2}{k}-1}$ on right side of my last equation then the expression on the right side of my first equation can be obtained (using Eq 3.194 of Gradeshteyn book (also given below)). However, I do not know where I am making a mistake and how to show that the relation in my first equation is true analytically. Please help me in getting it right. Many many thanks in advance.

Edit: After the steps of Harry Peter

I am very thankful to Harry Peter for his help (see the Hint in the answer below). Here, I extend the steps of Harry Peter further. The last step of Harry Peter is $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx$$ now we can write it as $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}dx-\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}x^{-m}dx$$ which can be further written as $$\frac{1}{ak}\left(\frac{-k}{2}\right)(a)\left[\frac{x}{a}-1\right]^{-\frac{2}{k}}\bigg{|}_1^{\frac{a}{u^k}+1}-\frac{(-1)^{-\frac{2}{k}-1}}{ak}\int_1^{\frac{a}{u^k}+1}\left(1-\frac{x}{a}\right)x^{-m}dx$$ putting limits in the first part and using substitution $z=\frac{x}{a}$ in the second part we can write $$=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-\frac{1}{ak}(-1)^{-\frac{2}{k}-1}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}(az)^{-m}a \, dz$$ which after simplification can be written as $$=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-(-1)^{-\frac{2}{k}-1}a^{-m}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz$$ $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{\frac{2}{k}}\right]-C$$ where $C$ is given as follows $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz$$ $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[\int_0^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz-\int_0^{\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz\right]$$ Now using the definition of incomplete Beta function ($B_z(p,q)=\int_0^zt^{p-1}(1-t)^{q-1}\,dt$) we can write $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[B_{\frac{1}{u^k}+\frac{1}{a}}(1-m,-\frac{2}{k})-B_{\frac{1}{a}}(1-m,-\frac{2}{k})\right]$$ Now using the definition of Hypergeometric function ($B_z(p,q)=\frac{z^p}{p}\, _2F_1(p,1-q;p+1;z)$)we can write $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[\frac{\left(\frac{1}{u^k}+\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{u^k}+\frac{1}{a})-\frac{\left(\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{a})\right]$$ Putting the value of $C$ in the above equation for $\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx$ we can write $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-(-1)^{-\frac{2}{k}-1}a^{-m}\left[\frac{\left(\frac{1}{u^k}+\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{u^k}+\frac{1}{a})-\frac{\left(\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{a})\right]$$ Now the right side is not looking the same as the right side of my very first equation of this post. Maybe there is some very nice property or identity of Hypergeometric function that can be used to make it look like the right side of very first equation. But unfortunately I am unaware of any such formula. I request you to help me in getting the right side of very first equation. Many many thanks in advance.

Thank you

• This does not seem to work : try $a=1,k=1,m=1,u=1$ and you integrate $\frac{x}{x+1}$ which does not converge. Changing $m=2$, you integrate $\frac{x (2 x+1)}{(x+1)^2}$ which does not converge either. – Claude Leibovici Oct 5 '16 at 9:07
• @ClaudeLeibovici Does your conclusion change if we make $k$ to be greater than $2$? – Frank Moses Oct 11 '16 at 9:16
• @ Frank Moses. Totally ! I tried $a=1,k=3,m=1$ and it works for all $u>0$. If fact, even $u=0$ works. – Claude Leibovici Oct 11 '16 at 9:33
• @ClaudeLeibovici so this means the formula presented in my first equation is right? But if this is right then why in Gradeshteyn book they mentioned that $Re(\mu)$ should be greater than zero. In my edited question I have added the image from the Gradeshteyn book also. Many thanks for your help. – Frank Moses Oct 11 '16 at 9:42
• In Gradshteyn's formula, if $\mathrm{Re}(\mu)\leq0$, the function is not integrable (look at what happens at $x=0$). However, your integral is slightly different (the big square bracket vanishes at $v=0$. – Jean-Claude Arbaut Oct 12 '16 at 12:57

Hint: $$\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x$$

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

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

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

• Many thanks for your answer. I have added the steps in my post (using the definition that you provided of incomplete Beta function and Hypergeometric function). But the problem is that the right side of my last expression does not look to be same as the right side of my very first eqauation in my post. Maybe there is some nice property which can be used to show that both are equal but I am unaware of such property. I would be very thankful to you if you could help in this regard. Thanks in advance – Frank Moses Oct 13 '16 at 2:27

Hint:

$\int_u^\infty\left(1-\dfrac{1}{\left(1+\dfrac{a}{x^k}\right)^m}\right)x~dx$

$=\int_\frac{1}{u}^0\left(1-\dfrac{1}{(1+ax^k)^m}\right)\dfrac{1}{x}~d\left(\dfrac{1}{x}\right)$

$=\int_0^\frac{1}{u}x^{-3}(1-(1+ax^k)^{-m})~dx$

$=\int_0^\frac{1}{u^k}x^{-\frac{3}{k}}(1-(1+ax)^{-m})~d\left(x^\frac{1}{k}\right)$

$=\dfrac{1}{k}\int_0^\frac{1}{u^k}x^{-\frac{2}{k}-1}(1-(1+ax)^{-m})~dx$

$=\dfrac{1}{k}\int_1^{\frac{a}{u^k}+1}\left(\dfrac{x-1}{a}\right)^{-\frac{2}{k}-1}(1-x^{-m})~d\left(\dfrac{x-1}{a}\right)$

$=\dfrac{a^\frac{2}{k}}{k}\int_1^{\frac{a}{u^k}+1}(x-1)^{-\frac{2}{k}-1}(1-x^{-m})~dx$

• What are next steps? I mean what substitution is to be used and after using substitution what integration identity should be used? Because with out substitution the first part of the integration does not converge. Many thanks in advance. – Frank Moses Oct 9 '16 at 0:29
• Maybe its very easy after these steps but unfortunately I have spent a lot of time in figuring out the way forward after these steps without having any success. Please add some more steps so that I can understand it. Thanks in advance – Frank Moses Oct 12 '16 at 3:37
• Many thanks Harry Peter. In my edited post, I have added the steps that I can do after the simplification that you provided. Now the right side is not looking the same as the right side of my very first equation of this post. Maybe there is some very nice property or identity of Hypergeometric function that can be used to make it look like the right side of very first equation. But unfortunately I am unaware of any such formula. I request you to help me in getting the right side of very first equation. Many many thanks in advance. – Frank Moses Oct 13 '16 at 2:12