How to show that $\int_0^x \left(\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}\right)(x-t)^n\mathrm dt\to0$ when $n\to\infty$, for $\vert x\vert < 1$?

Question

I want to show that : $$ \forall \vert x\vert < 1 \:,\: \int_0^x \left(\dfrac{1}{(t-1)^{n+1}}-\dfrac{1}{(t+1)^{n+1}}\right)(x-t)^n\mathrm dt \xrightarrow[n\to\infty]{} 0 $$ I've tried to bound from above $(x-t) ^n$ but $\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}$ seems to tend towards $\infty$. I've also tried with absolute value (hope to find an upper bound) but my attemps failed. I don't see how to show that this integral tends to $0$.

Answer 1

Split the integral into the two parts

$$ \int_0^x \left(\frac{x-t}{t-1}\right)^n (t-1)^{-1}\,dt - \int_0^x \left(\frac{x-t}{t+1}\right)^n (t+1)^{-1}\,dt =: I_1(n,x) - I_2(n,x). $$

For simplicity we'll suppose that $x > 0$ and leave the $x<0$ case out of this answer. So for us $0 < x < 1$.

We can rewrite

$$ I_1(n,x) = (-1)^{n+1} \int_0^x \left(\frac{x-t}{1-t}\right)^n (1-t)^{-1}\,dt. $$

Now

$$ \frac{d}{dt} \frac{x-t}{1-t} = \frac{x-1}{(1-t)^2} < 0 $$

for $0 \leq t \leq x < 1$, so

$$ 0 \leq \frac{x-t}{1-t} \leq \left.\frac{x-t}{1-t}\right|_{t=0} = x, $$

and thus

$$ 0 \leq \int_0^x \left(\frac{x-t}{1-t}\right)^n (1-t)^{-1}\,dt \leq x^n \int_0^x (1-t)^{-1}\,dt = -x^n \log(1-x). $$

It follows that

$$ |I_1(n,x)| \leq -x^n \log(1-x). \tag{$*$} $$

A similar argument can be used for $I_2(n,x)$. Since

$$ \frac{d}{dt} \frac{x-t}{t+1} = - \frac{x+1}{(t+1)^2} < 0 $$

we have

$$ 0 \leq \frac{x-t}{t+1} \leq x, $$

and thus

$$ 0 \leq I_2(n,x) \leq x^n \log(1+x). \tag{$**$} $$

Combining $(*)$ and $(**)$ yields

$$ \begin{align*} \left| \int_0^x \left(\frac{1}{(t-1)^{n+1}}-\frac{1}{(t+1)^{n+1}}\right)(x-t)^n\,dt \right| &\leq x^n [\log(1+x)-\log(1-x)] \\ &= x^n \log\!\left(\frac{1+x}{1-x}\right). \tag{$***$} \end{align*} $$

Thus your integral tends to $0$ as $n \to \infty$.

Answer 2

Here is a quick and dirty way of getting the answer. Everything I am sure can be rigorously justified.

It is well known that (I believe as the Cauchy repeated integral formula) $$\int_0^x f(t)(x-t)^n dt = I_0^n f(x)= \int_0^x \int_0^x \cdots \int_0^x f(t) dt,$$ where if you excuse the sloppiness, we have repeated integration with the same lower bound of 0.

EDIT: Not confident in my calculations below. If done right it should work. I believe the second term is wrong for each integration but they should both be equal.

The $n$th integral of $1/(x-1)^{n+1}$ is $(-1)^n/n! \cdot 1/(x-1)+(-1)^{n+1}$.

Similarly, for $1/(x+1)^{n+1}$ it is $(-1)^n/n! \cdot 1/(x+1) + (-1)^{n+1}$.

Then the limit is easily seen to be zero.