# $\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$

Define $$f(x)=\dfrac{x+1}{(x-1)^2}$$. Prove $$\lim\limits_{x \to 1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, {\rm d}t=\dfrac{\pi }{\sqrt{2}}$$.

We can obtain $$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$ but how to go on ?

• $$\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}=\frac{\sqrt{t}(t+1)}{(t-1)^2\sqrt{f(x)-\frac{t+1}{(t-1)^2}}}=\frac{\sqrt{t}(t+1)}{(t-1)\sqrt{f(x)(t-1)^2-(t+1)}}$$ Commented Dec 7, 2020 at 18:45
• @HenryLee Does it help? Commented Dec 7, 2020 at 19:50

I found a possible method here which is straightforward. I will just carry it here.

\begin{aligned} \lim _{x \rightarrow 1} u(x) &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t} \cdot \frac{1+t}{(t-1)^{2}}}{\sqrt{\frac{1+x}{(x-1)^{2}}-\frac{1+t}{(t-1)^{2}}}} d t \\ &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t}(1+t)(1-x)}{(1-t) \sqrt{(x-t)(3-t-x-x t)}} d t \\ &=\lim _{y \rightarrow 0} \int_{y}^{1} \frac{(2-s) \sqrt{1-s} y}{s \sqrt{(s-y)(2 s+2 y-s y)}} d s \quad(\text { Let } s=1-t, y=1-x) \\ &=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y} \cdot y \cdot y}{u y \sqrt{(u y-y)\left(2 u y+2 y-2 u y^{2}\right)}} d u \quad(\text { Let } s=u y) \\ &=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y}}{u \sqrt{(u-1)(2 u+2-u y)}} d u\\ &=\sqrt{2} \int_{1}^{+\infty} \frac{1}{u \sqrt{u^{2}-1}} d u \\ &=\sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-m^{2}}} d m \quad\left(\text { Let } m=\frac{1}{u}\right) \\ &=\frac{\sqrt{2}}{2} \pi \end{aligned}

I'm somehow not really satisfied with all those great solutions above. To me, they didn't show quite well the underlying reason that makes that convergence occur. So I'll give another solution by trying to generalize the very first problem.

I'll start by stating something trivial.

Lemma 1:
(Convergence of measures)
Let $$\mu_1, \mu_2,...$$ be a sequence of measures on $$[0,1]$$ such that $$\lim_{ n \rightarrow \infty} \mu_n[0,1-\epsilon) = 0$$ for all $$\epsilon>0$$, and $$\lim_n \mu_n [0,1]=C$$ for some $$C>0$$, then for all continuous function $$g$$ on $$[0,1]$$ , we have: $$\int g d\mu_n \longrightarrow Cg(1)$$ $$\square$$

Now, let's go back to the initial question.
Main solution

I'll define:

• $$g(t):= \frac{ \sqrt{2t} f(t)^{3/2}}{f'(t)}$$
• $$d\mu_x(t)= \mathbb{1}_{[0,x]}(t)\frac{1}{ \sqrt{\left(1- h_x(t)\right)h_x(t)}} h_x'(t)dt$$, where $$h_x(t)= \frac{f(t)}{f(x)}$$

So we see that:

• The integral whose limit we want to calculate is equal to $$\int g d\mu_x$$
• $$g$$ is continuous on $$[0,1]$$ and $$g(1)=\frac{\sqrt{2}}{2}$$
• Note that $$h_x$$ is increasing in $$t$$, thus by changing variables, we can show that $$\mu_x[0,1] = \int_{[0,1]} \frac{1}{\sqrt{u(1-u)}} du= \pi$$
• Also by changing variables and noting that $$\lim_{x \rightarrow 1^-} f(x)=+\infty$$, we can show that: $$\lim_{x \rightarrow 1^{-}} \mu_x[0,\alpha] =0$$ for all $$\alpha \in (0,1)$$

Hence by applying the very first lemma, we have the final result

Comment:
TL;DR: the main idea is just to find the right $$g$$ and $$\mu_x$$ to compactify our integral.

Finally I was able to handle this problem. It is just tedious calculation. Let $$u=\frac{f(t)}{f(x)}$$ and then $$\begin{eqnarray}t&=&1-\frac{(1-x)(x-1+\sqrt{(1-x)^2+8u(1+x)}}{2u(1+x)},\\ dt&=&-\frac12\frac{ (x-1) \left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)}{u^2(1+x)\sqrt{(1-x)^2+8u(1+x)}}du. \end{eqnarray}$$ So $$\begin{eqnarray} &&\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t\\ &=&\lim_{x \to 1}\int_{\frac1{f(x)}}^1\frac{\sqrt{t}\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2 }{u \sqrt{(1-u) (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}{\rm d}u\\ &=&\int_0^1\frac{1}{\sqrt{2u(1-u)}}du\\ &=&\frac{\pi}{\sqrt 2}. \end{eqnarray}$$ Here $$\lim_{x \to 1}\frac1{f(x)}=0, \lim_{x \to 1}t=1,$$ and it turns out that when $$x=1$$, the integrand $$\frac{\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2 \sqrt{\frac{(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)}{2 u (x+1)}+1}}{u \sqrt{(1-u) (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}\bigg|_{x=1}=\frac{1}{\sqrt{2u(1-u)}}$$

• Out of curiosity, how can you let $\lim_{x \rightarrow 1}$ enter inside the intergral? Commented Dec 9, 2020 at 0:22
• It needs some work for the limit to go inside the integral. Commented Dec 9, 2020 at 0:25
• hm, dear Paul, is there any concise argument to make me understand it though? Or it requires more calculations to have that part clear? Commented Dec 9, 2020 at 0:28
• Actually, it turns out that when $x=1$, the integrand is $\frac{1}{\sqrt{2u(1-u)}}$.. Commented Dec 9, 2020 at 0:29
• I think is easy to handle this issue. When I have time, I will fix it. Commented Dec 9, 2020 at 0:48