# Is $\int_{-1}^1 \frac{dx}{\sqrt{x^2 - 1}}$ divergent?

I would like to know if the following integral is divergent:

$$\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}} = \pi$$

Wolfram alpha returned a finite answer of $\pi$. It looks like it should have poles at $x=-1,1$. Can explain?

The antiderviative is $\int dx \,(1-x^2)^{-1/2} = \sin^{-1} x$ but I want to know why the divergence goes away.

Edit: An earlier version had this one:

$$\int_{-1}^1 \frac{dx}{x^2 - 1}$$

Certainly it has poles on both ends. Maybe we can use the partial fraction decomposition:

$$\frac{2}{x^2 -1} = \frac{1}{x-1} - \frac{1}{x+1}$$

• Not only poles, but bad poles. The (improper) integral does not exist. Commented Sep 15, 2014 at 22:39
• For your completely changed problem, we have convergence, except that of course the function is not real-valued. Probably something other than the current title is intended, involving $\sqrt{1-x^2}$. The integral $\int_{-1}^1 \frac{1}{\sqrt{1-x^2}}\,dx$ converges. Commented Sep 15, 2014 at 22:47
• @AndréNicolas yes but why does it converge despite integrand tending to infinity? I think it's like $$\int_0^1 \frac{dx}{\sqrt{x}} = \sqrt{x}\bigg|_0^1 = 1$$ Commented Sep 15, 2014 at 22:49
• Basically it is for the same reason as the integral of $1/\sqrt{x}$. Someone will likely write out a formal comparison argument. If no one else has, I may when I return from the Department. Commented Sep 15, 2014 at 22:52
• Your title says $\sqrt{x^2-1}$ whereas your question says $\sqrt{1-x^2}$ ... I answered to both to make sure you get your answer Commented Sep 15, 2014 at 23:03

## 3 Answers

$$\int_{-1}^1 \frac{dx}{x^2 - 1}=\int_{-1}^1 \left(\frac{1}{2 x - 2}- \frac{1}{2 x + 2}\right) \, dx$$

$$\int_{-1}^1\frac{1}{x + 1}=[\ln(x+1)]_{-1}^1.$$

However, $\ln(x+1)$ isn't defined for $x=-1$, and $\displaystyle\lim_{x\rightarrow-1}\ln(x+1)=-\infty$.

Therefore $\int_{-1}^1 \frac{dx}{x^2 - 1}$ is not defined.

$$\int_{-1}^1 \frac{dx}{\sqrt{x^2 - 1}}=[\operatorname{acosh}(x)]_{-1}^1=-i\pi$$

This is well known because the derivative of $\operatorname{acosh}(x)$ is $\frac{1}{\sqrt{x^{2} - 1}}$

$$\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}}=[\operatorname{asin}(x)]_{-1}^1=\pi$$

This is well known because the derivative of $\operatorname{asin}(x)$ is $\frac{1}{\sqrt{1-x^2}}$

• up vote --- sorry for switching it up on you Commented Sep 15, 2014 at 22:41

We want $$\lim_{(\delta,\epsilon)\to(0^+,0^+)}\int_{-1+\delta}^{1-\epsilon} \frac{1}{\sqrt{1-x^2}}\,dx.$$ This is $$\lim_{(\delta,\epsilon)\to(0^+,0^+)}\left(\arcsin(1-\epsilon)-\arcsin(-1+\delta)\right).$$ But $$\lim_{\epsilon\to 0^+}\arcsin(1-\epsilon)=\frac{\pi}{2}\quad\text{and}\quad \lim_{\delta\to 0^+}\arcsin(-1+\delta)=-\frac{\pi}{2}.$$

Remark: If we are only interested in existence (convergence), it is convenient to break up the integral into two parts, in order to have only one element of "badness" at a time. We show for example that $$\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx\tag{1}$$ exists. Note that on our interval the function is positive and les than or equal to $\frac{1}{\sqrt{1-x}}$, since $1+x\ge 1$. If we can show that $$\int_0^1 \frac{1}{\sqrt{1-x}}\,dx\tag{2}$$ exists, then by Comparison so does the integral (1). To show that the integral (2) exists, integrate $\frac{1}{\sqrt{1-x}}$ explicitly from $0$ to $1-\epsilon$, and take the limit as $\epsilon\to 0$ from the right.

Or else, take the "badness" to $0$ (my preference) by making the change of variable $1-x=t$, and refer to the standard fact that $\int_0^1 \frac{1}{\sqrt{t}}\,dt$ exists.

Hint: Notice that, due to the parity of the integrand, our expression becomes $I=2\displaystyle\int_0^1\dfrac{dx}{\sqrt{1-x^2}}$.

Now, let $x=1-t$. Our integral becomes $I=2\displaystyle\int_0^1\dfrac{dt}{\sqrt{t~(2-t)}}$, whose convergence is the same

as that of $J=\displaystyle\int_0^1\dfrac{du}{\sqrt u}$, since the term $\dfrac1{\sqrt{2-t}}$ causes no trouble on $[0,1]$.