2

I have difficulties with solving the convergence of Newton inetgral: $$\int_0^1\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}, a\in\mathbb{R}$$

Edit Show for which $a$ does it converges.

I don't know much where/how to start so I would appreciate any help.

My suggestions:

Divide it on two intervals, let's say $$\int_0^{1/2}\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a} +\int_{1/2}^1\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}$$

In the first interval, using limit comparison test with function $g(x)=\frac{1}{\sqrt{x}}$, we get $$lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\cdots=0$$ And since $\int_0^{1/2}\frac{1}{\sqrt{x}}$ converges, we have, that$\int_0^{1/2}\log{(\arctan{}x)}\dfrac{\frac{\pi}{2}-\arcsin{x}}{(e^{1-x}-1)^a}$ converges for all $a$.

But the second part seems to be trickier.

Edit 2:

Is it possible that we for the second part we will use limit comparison test with function $g(x)=(x-1)^{-a+1}$? Then we will get that it converges for all $a<1$...but it might be wrong

GorTeX
  • 553

2 Answers2

2

Let $I(a)$ be the integral

$$I(a)=\int_0^1\,\log(\arctan x)\frac{\pi/2-\arcsin x}{(e^{1-x}-1)^a} \tag1$$

where $a$ is a real-valued parameter.

Note that the integrand of $(1)$ has singularities at both $x=0$ and $x=1$. We shall investigate these singularities to determine when the integral in $(1)$ exists as an improper Riemann integral.

First, the singularity at $x=0$ arises in the term $\log(\arctan x)$. Noting that $\arctan x =x+O(x^3)$, we have

$$\begin{align} \log \arctan x &=\log (x+O(x^3))\\\\ &=\log x+\log(1+O(x^2))\\\\ &=\log x+O(x^2) \end{align}$$

Inasmuch as $\int_0^1 \log x$ exists as an improper Riemann integral, we conclude that the singularity at $x=0$ does not render the integral divergent. However, we still need to check the singularity at $x=1$.

The singularity at $x=1$ arises for $a>0$ due to the term $\frac{1}{(e^{1-x}-1)^a}$. This singularity is of order $(1-x)^{-a}$ for $a>0$. More precisely,

$$\frac{1}{(e^{1-x}-1)^a}=(1-x)^{-a}(1+O(1-x))^{-a}$$

We also have (see NOTE)

$$\pi/2-\arcsin x=\sqrt{2}\sqrt{1-x}+O(1-x)^{3/2} \tag 2$$

so that

$$\frac{\pi/2 -\arcsin x}{(e^{1-x}-1)^a}=\left(\sqrt{2}\sqrt{1-x}+O(1-x)^{3/2}\right)\left((1-x)^{-a}[1+O(x-1)]^{-a}\right)\tag 3$$

The leading term of $(3)$ is of order $(1-x)^{1/2-a}$. Thus, the singularity at $x=1$ will lead to a convergent improper Riemann integral for values $a<3/2$. Otherwise the integral diverges.

NOTE: On the Expansion of $\arcsin(x)$ Around $x=1^{-}$.

Let $f(x)=\arcsin(x)$. Then, $f'(x)=\frac{1}{\sqrt{1-x^2}}$. Therefore, $f'(x)\sim \frac{\sqrt 2}{2}\frac{1}{\sqrt{1-x}}$ near $x=1^{-}$.

This motivates enforcing the substitution $x=1-y^2$ so that $y=\sqrt{1-x}$. Note that as as $x\to 1^{-}$, $y\to 0^{+}$.

We proceed now to expand $f(x)=f(1-y^2)=\arcsin(1-y^2)$ in a series around $y=0$. Taking the first few derivatives of $f(1-y^2)$ with respect to $y$ gives for $y\ge 0$

$$\begin{align} \frac{df}{dy}&=-2\left(2-y^2\right)^{-1/2}\\\\ \frac{d^2f}{dy^2}&=-2y\left(2-y^2\right)^{-3/2}\\\\ \frac{d^3f}{dy^3}&=-4(y^2+1)\left(2-y^2\right)^{-5/2}\\\\ \end{align}$$

Evaluating these derivatives at $y=0^+$ provides the expansion coefficients for the expansion of $\arcsin(1-y^2)$

$$\arcsin(1-y^2)=\frac{\pi}{2}-\sqrt 2 y-\frac{\sqrt 2}{12}y^3+O(y^5)$$

and therefore we have

$$\arcsin(x)=\frac{\pi}{2}-\sqrt 2 \sqrt{1-x}\left(1+\frac12 (1-x)+O(1-x)^2\right)$$

Mark Viola
  • 179,405
  • This seems nice to me. In general, this decading, wheteher an integral converges or diverges, seems to me quite hard - in this case I have made an error that I evaluated $\pi/2-\arcsin x$ as similar to $1-x$ thus I have arrived to the result $a<1$. How did you get to the evaluation of $\pi/2-\arcsin x$? Thx – GorTeX Jun 04 '15 at 19:28
  • @GorteX Thank you! To answer your question, expand $f(y)=\arcsin (1-y^2)$ in a Taylor series around $y=0$, then set $y=\sqrt{1-x}$. The result is an expansion of $\arcsin(x)$ around $x=1$. – Mark Viola Jun 04 '15 at 19:43
  • I dont get it- why does it lead to expansion around $x=1$? T – GorTeX Jun 04 '15 at 19:50
  • I see it a bit but still I have problem with the second term - the second term of Taylor series in our case should be: $f'(0)(x)$. The derivative is $f'(y)=\frac{-2y}{\sqrt{1-(1-y^2)^2}}$ and when I put $y=0$, then I divide by 0. – GorTeX Jun 04 '15 at 20:02
  • @gortex You're absolutely correct. Only the odd terms survive ($(1-x)^{1/2}$,$(1-x)^{3/2}$, etc.). For your concern, $\sqrt{1-(1-y^2)^2}=y\sqrt{2-y^2}$. – Mark Viola Jun 04 '15 at 20:07
  • Nice...it bothers me that I cannot get the expansion at 1 from the orginal function $\arcsin x$. How did you come up with this nice trick? :-) – GorTeX Jun 04 '15 at 20:15
  • @gortex Well, we know that the arcine has patholgies in its derivatives around $1$. So, we have to analyze this behavior. We know that its first derivative is $(1-x^2)^{-1/2}\sim \sqrt{2}\sqrt{1-x}$ for $x$ "close" to $1$. – Mark Viola Jun 04 '15 at 20:24
  • Ok, thank you for your help. I really appreciate it... – GorTeX Jun 04 '15 at 20:26
  • @gortex You're welcome! My pleasure. – Mark Viola Jun 04 '15 at 20:40
  • @Dr.MV, could you comment a little bit more on your expansion of ${\rm arcsin}\ x$ around $x = 1$? Is your approach purely "syntactic" or is there some analysis I don't get? I thought it wouldn't be possible to do the expansion at all. – David Sep 13 '15 at 14:09
  • @DavidČepelík Please see the previous comments in this post. We can expand $\arcsin(1-y^2)$ in a Taylor series around $y=0$. Then, simply let $y=\sqrt{1-x}$. This results in a series around $x=1$ for $\arcsin (1-(\sqrt{1-x})^2)=\arcsin(x)$. – Mark Viola Sep 13 '15 at 14:24
  • I did see them and I tried. (Thanks, by the way, it's a really clever way to get around.) What I mean is; how can it be that we avoid the problems with infinite derivatives of ${\rm arcsin}\ x$ by substituting back and forth? From the definition of Taylor polynomial, I would not even guess this is remotely possible and would probably think that such growth just cannot be approximated by a polynomial. – David Sep 13 '15 at 14:26
  • Thinking about it, it's probably the truth, as what we get is not a Taylor polynomial, right? We just use the TP of a little bit different function to get somewhere and then change the argument to finally arrive at an approximation, which works in this case, but it's not polynomial. Am I correct? – David Sep 13 '15 at 14:29
  • @DavidČepelík See if you can expand $\frac{\pi/2-\arcsin(x)}{\sqrt{1-x}}$ around $x=1$. The expansion will be a Taylor series in powers of $(x-1)$. – Mark Viola Sep 13 '15 at 14:29
  • @Dr.MV OK, I'll try that! – David Sep 13 '15 at 14:30
1

For the second interval, note first the $\log $ term has a finite nonzero limit at $1,$ so for matters of convergence we can ignore it. Second, recall $(e^h-1)/h\to 1$ as $h\to 0.$ This is just the definition of the derivative of $e^x$ at $0.$ It follows that in absolute value the denominator is on the order of $(1-x)^a.$ Upstairs we have

$$\frac{\pi}{2}- \arcsin x = \int_x^{1} \frac{dt}{\sqrt {1-t^2}}.$$

Now $\sqrt {1-t^2}$ is on the order of $\sqrt {1-t}$ if $t$ is near $1.$ You can integrate $1/\sqrt {1-t}$ easily to see the last integral is on the order of $(1-x)^{1/2}.$ It looks to me like your second integral converges iff $1/2-a>-1.$

zhw.
  • 105,693