Show that $\lim_{x \to 0}\int_0^1 \frac{t^x}{(1+t^x)^2}dt = \frac{1}{4}$

Question

I have to show that $$\lim\limits_{x \to 0} \displaystyle\int_0^1 \dfrac{t^x}{(1+t^x)^2}dt = \dfrac{1}{4} $$

I tried to show that $$\displaystyle\int_0^1 \dfrac{t^x}{(1+t^x)^2} - \dfrac{1}{4} dt= \displaystyle\int_0^1 \left(\dfrac{t^x-1}{t^x+1}\right)^2 dt \rightarrow 0$$ when $x \rightarrow 0$.

And to show that I tried to bound $\displaystyle\int_0^1 \left(\dfrac{t^x-1}{t^x+1}\right)^2 dt$ but it didn't work.

Then I substitute $u=t^x$ and it gave me that $$\displaystyle\int_0^1 \dfrac{t^x}{(1+t^x)^2}dt = \displaystyle\int_0^1 \dfrac{u^{\frac{1}{x}}}{x(1+u)^2}du$$ but I don't see how to simplify that.

Thanks for your help!

Answer 1

Let $x>0$. Noting $$ \left(\dfrac{1-t^x}{1+t^x}\right)^2=\dfrac{1-t^x}{(1+t^x)^2}(1-t^x)\le 1-t^x$$ one has $$\bigg|\int_0^1 \dfrac{t^x}{(1+t^x)^2}dt - \dfrac{1}{4}\bigg|= \displaystyle\int_0^1 \left(\dfrac{1-t^x}{1+t^x}\right)^2 dt\le\int_0^1(1-t^x)dt=\frac x{1+x}\rightarrow 0$$ as $x\to0$.

Answer 2

Since you're not familiar with the dominated convergence theorem, I'll approach this using elementary methods. Set

$$f_x(x)=\left(\frac{t^x-1}{t^x+1}\right)^2=\left(1-\frac{2}{t^x+1}\right)^2.$$

The goal is to show that

$$\lim_{x\to0}\int_0^1 f_x(t)~\mathrm{d}t=0.$$

I'll start by showing that

$$\lim_{x\to 0}\int_\delta^1 f_x(t)~\mathrm{d}t=0$$

for all $\delta\in(0,1)$. So fix $\delta\in(0,1)$. Then, for any $t\in[\delta,1]$ we have that

$$f'_x(t)\leq0$$

(check this for yourself), so $f_x$ is decreasing on $[\delta,1]$. In particular then

$$0\leq f_x(t)\leq f_x(\delta).$$

Consequently

$$0\leq \int_\delta^1 f_x(t)~\mathrm{d}t\leq \int_\delta^1f_x(\delta)~\mathrm{d}t=f_x(\delta)(1-\delta),$$

and so as

$$\lim_{x\to0}f_x(\delta)=\lim_{x\to0}\left(1-\frac{2}{\delta^x+1}\right)^2=0,$$

it follows that

$$\lim_{x\to 0}\int_\delta^1 f_x(t)~\mathrm{d}t=0.$$

Let us now use this limit to prove our result. Let $\varepsilon>0$. Observe that

$$0\leq f_x(t)\leq 1$$

for all $t\in[0,1]$ and all $x\in\mathbb{R}$, so

$$0\leq\int_0^{\frac{\varepsilon}{2}} f_x(t)~\mathrm{d}t\leq\int_0^{\frac{\varepsilon}{2}}\mathrm{d}t=\frac{\varepsilon}{2}.$$

Using the limit we proved before, we now find a $\xi>0$ such that

$$0\leq \int_{\frac{\varepsilon}{2}}^1 f_x(t)~\mathrm{d}t<\frac{\varepsilon}{2}$$

whenever $\lvert x\rvert<\xi$. It now follows that whenever $\lvert x\rvert<\xi$,

$$0\leq\int_0^1 f_x(t)~\mathrm{d}t=\int_0^{\frac{\varepsilon}{2}}f_x(t)~\mathrm{d}t+\int_{\frac{\varepsilon}{2}}^1f_x(t)~\mathrm{d}t<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

It follows that

$$\lim_{x\to0}\int_0^1f_x(t)~\mathrm{d}t=0,$$

from which your limit follows.

Answer 3

I thought it might be instructive to present a solution without appealing to the Dominated Convrgence Theorem. To that end, we proceed.

First, we note that

$$\int_0^1 \frac{(1-t^x)^2}{(1+t^x)^2}\,dt=\int_0^1 \frac{(1-t^{|x|})^2}{(1+t^{|x|})^2}\,dt$$

and therefore, without loss of generality, we can assume that $x>0$ and that the limit is taken from the right.

Now, for any given $\varepsilon \in (0,1]$, we have

$$\begin{align} \int_0^1 \frac{(1-t^x)^2}{(1+t^x)^2}\,dt&=\int_0^\varepsilon \frac{(1-t^{|x|})^2}{(1+t^{|x|})^2}\,dt+\int_\varepsilon^1 \frac{(1-t^{|x|})^2}{(1+t^{|x|})^2}\,dt\\\\&\le \varepsilon (1-\varepsilon^{|x|})^2+\int_\varepsilon^1 \frac{(1-t^{|x|})^2}{(1+t^{|x|})^2}\,dt \end{align}$$

Letting $x\to 0$, we find that

$$\begin{align} \lim_{x\to 0}\int_0^1 \frac{(1-t^x)^2}{(1+t^x)^2}\,dt&\le \varepsilon \end{align}$$

Inasmuch $0<\varepsilon\le 1$ is arbitrary, we conclude that

$$\lim_{x\to 0}\int_0^1 \frac{(1-t^x)^2}{(1+t^x)^2}\,dt=0$$

as we are done!

Answer 4

Hint

The simplest is probably to write $t^x=e^{x \log(t)}$ and to use the series expansion around $x=0$ and then integrate termwise.

More complex is to remember the integral representation of the harmonic numbers as given by Euler $$H_n=\int_0^1 \frac{1-z^n }{1-z }\,dz$$ to obtain the general result $$\frac{x+H_{\frac{1}{2} \left(\frac{1}{x}-1\right)}-H_{\frac{1}{2 x}}}{2 x^2}$$ Let $x=\frac 1y$ and use the asymptotics of the harmonic numbers.

Answer 5

It's clear that the limit of the integrand is $\frac{1}{4}$, so the integral itself is $\frac{1}{4}$.

However, we have to justify exchanging the limit and integral signs. It suffices to show that the integrand is uniformly convergent. In other words, to show that the integrand is differentiable and bounded, what it clearly is.

Answer 6

For $t\in (0,1)$ we have $f(x,t) := \frac{t^x}{(1+t^x)^2}\leqslant 1$ as well as $$ \lim _{x\to 0} \frac{t^x}{(1+t^x)^2} = \frac{1}{4}. $$ Hence by dominated convergence $$ \lim _{x\to 0} \int _0^1 f(x,t)dt = \int _0^1\lim _{x\to 0} f(x,t)dt = \int _0^1\frac{1}{4}dt = \frac{1}{4}. $$