Ideas on how to estimate this parameter integral Consider the following integral function where $V$ is the $d$-dimensional unit ball and $0<r<1$
$$f(r)=\int_V \frac{1}{|rx-\frac{1}{r}\mathbf{1}|}-\frac{1}{|x-\mathbf{1}|}dx,$$
where $\mathbf{1}=(1,1,\dots,1)\in \mathbb{R}^d$.
I want to find an upper bound which depends explicitely on $r$, i.e. something like $f(r) \leq rA+B$ where $A,B$ are just constants. It is enough for me to find an upper bound in the case where $r$ is close to $1$.
So my idea is to use a Taylor expansion around $r_o=1$ for the first part of the integrand $g:\mathbb{R}\rightarrow\mathbb{R}, g(r):=\frac{1}{|rx-\frac{1}{r}\mathbf{1}|}.$
Then $$g(1)=\frac{1}{|x-\mathbf{1}|}$$
which will cancel the other term in the integral and
$$g'(1)=\frac{d^2-\Big(\sum_{i=1}^d x_i-1\Big)^2}{|x-\mathbf{1}|^3}.$$
if I made no mistake in differentiating.
So by Taylors formula (and cancelling out the constant terms, splitting the integrals and pulling the $r$ out of the integral)
$$f(r)=r\underbrace{\int_V \frac{d^2-\Big(\sum_{i=1}^d x_i-1\Big)^2}{|x-\mathbf{1}|^3}dx}_{=:A}  +\underbrace{\int_VR_2(r,x)dx}_{=:B}.$$
So I would need to estimate the error term $R_2(r,x)$ but this will be very painful since $g''$ will look awful.
My questions: What do you think about this approach? Does there exist an easier approach? Do you think this way will give me an upper bound? If I use Taylor, do you see an easy way to calculate $g''$ maybe by finding an easier expression for $g$ or $g'$? Additionally, $f''$ will have a singularity as well, so I won't get any upper bounds?! So what can I do?
I appreciate any comment and idea.
 A: Let me summarize what has been commented so far:
By spherical symmetry, we can rewrite the integral:
$$f(r)=\int_V \frac{1}{|rx-\frac{1}{r}\mathbf{1}|}-\frac{1}{|x-\mathbf{1}|}dx=\int_V \frac{1}{|rx-\frac{1}{r}x_0|}-\frac{1}{|x-x_0|}dx,$$
where $x_0 = (\sqrt{d}, 0,\dots, 0)$. 
We also decompose $x = (t,y)$ such that $V =\{x \in \mathbb{R}^d \mid |x| <1 \}=\{(t,y) \in \mathbb{R} \times \mathbb{R}^{d-1} \mid -1<t<1, |y| < \sqrt{1-t^2}\}$ and therefore write 
$$f(r) = \int_{-1}^1 \int_{|y|<\sqrt{1-t^2}} \frac{1}{|r(t,y)-\frac{1}{r}x_0|}-\frac{1}{|(t,y)-x_0|}dydt$$
which equals
$$\int_{-1}^1 \int_{|y|<\sqrt{1-t^2}} \frac{1}{\sqrt{(rt-\frac{1}{r}\sqrt{d})^2+(ry)^2)}}-\frac{1}{\sqrt{(t-\sqrt{d})^2+y^2}}dydt,$$
where $y$ is always $d-1$-dimensional, so all products and squares are in scalar product sense.
However, is a Taylor expansion now the right way to get an upper bound which is good for $r$ close to $1$?
A: For $d = 1$, the integral diverges for any $r$. 
For $d \geq 2$, we can do some simple algebraic manipulations to get bounds on $f(r)$: first rewrite this as 
$$f(r) 
= \int_V \frac{|x - \mathbf{1}| - |rx - \frac{1}{r} \mathbf{1}|}{|x - \mathbf{1}||rx - \frac{1}{r} \mathbf{1}|} \,dx 
= \int_V \frac{|x - \mathbf{1}|^2 - |rx - \frac{1}{r} \mathbf{1}|^2}{|x - \mathbf{1}||rx - \frac{1}{r} \mathbf{1}|\left(|x - \mathbf{1}| + |rx - \frac{1}{r} \mathbf{1}|\right)} \,dx$$
and note that $|x - \mathbf{1}|^2 = |x|^2 + d - 2\langle x, \mathbf{1}\rangle$ and $|rx - \frac{1}{r}\mathbf{1}|^2 = r^2 |x|^2 + \frac{d}{r^2} - 2\langle x, \mathbf{1}\rangle$, so we have
$$f(r) = \int_V \frac{-(1-r^2)(\frac{d}{r^2} - |x|^2)}{|x - \mathbf{1}||rx - \frac{1}{r} \mathbf{1}|\left(|x - \mathbf{1}| + |rx - \frac{1}{r} \mathbf{1}|\right)} \,dx.$$
Now we can bound separately each of the factors in the integrand: we have $\frac{d}{r^2} - 1 \leq \frac{d}{r^2} - |x|^2 \leq \frac{d}{r^2}$, $\sqrt{d}-1 \leq |x - \mathbf{1}| \leq \sqrt{d} + 1$, and $\frac{\sqrt{d}}{r} - r \leq |rx - \frac{1}{r} \mathbf{1}| \leq \frac{\sqrt{d}}{r} + r$, hence we can bound our integral as
$$\frac{(1-r^2)(\frac{d}{r^2}-1)}{(\sqrt{d}+1)(\frac{\sqrt{d}}{r}+r)(\sqrt{d}+1+\frac{\sqrt{d}}{r}+r)}|V|\leq -f(r) \leq \frac{(1-r^2)\frac{d}{r^2}}{(\sqrt{d}-1)(\frac{\sqrt{d}}{r}-r)(\sqrt{d}-1 + \frac{\sqrt{d}}{r}-r)}|V|$$
where $|V|$ is the volume of $V$. After cancelling factors, this can be rewritten in the following form
$$\frac{\sqrt{d}-r}{(\sqrt{d}+1)(\sqrt{d}+r^2)}|V|\leq -\frac{f(r)}{1-r} \leq \frac{d}{(\sqrt{d}-1)(\sqrt{d}-r^2)(\sqrt{d}-r)}|V|.$$
and to get cleaner bounds near $r = 1$, note that the LHS is decreasing in $r$, while the RHS is increasing in $r$, hence taking their values at $r = 1$, we have $\frac{\sqrt{d}-1}{(\sqrt{d}+1)^2}|V| \leq -\frac{f(r)}{1-r} \leq \frac{d}{(\sqrt{d}-1)^3}|V|$, which we can write as 
$$-\frac{d|V|}{(\sqrt{d}-1)^3}(1-r) \leq f(r) \leq -\frac{(\sqrt{d}-1)|V|}{(\sqrt{d}+1)^2}(1-r).$$
