# Determine whether the improper integral $\int_{1}^{\infty} \frac{1}{x^3\sqrt{x^2 -1}} dx$ converges

Is this a proper proof?

Since the integrand exhibits a singularity at $$x=\pm1$$, and $$x=1$$ belongs to our domain of integration, we need to separate the integral into two summands:

$$\int_{1}^{\infty} \frac{1}{x^3\sqrt{x^2 -1}} dx = \int_{1}^{2}\frac{1}{x^3\sqrt{x^2 -1}} dx + \int_{2}^{\infty} \frac{1}{x^3\sqrt{x^2 -1}} dx$$

The first one converges:

$$\int_{1}^{2} \frac{1}{x^3(x+1)^{1/2}(x-1)^{1/2}} dx \stackrel{u=x-1}{=} \int_{0}^1 \frac{1}{(u+1)^{3} (u+2)^{1/2} u^{1/2}} dx$$

Given that $$\int_0^1 \frac{1}{x^{1/2}} dx$$ converges, applying the Limit Criterion...

$$\lim_{u\to 0^+} \frac{\frac{1}{(u+1)^{3} (u+2)^{1/2} u^{1/2}}}{\frac{1}{u^{1/2}}} = \lim_{u\to 0^+} \frac{1}{(u+1)^3(u+2)^{1/2}} = \frac{\sqrt{2}}{2} \neq 0$$

...we find that our first integral also converges.

The second one converges:

For $$x \in [2,+\infty)$$ we have that

$$\frac{1}{x^3 (x^2-1)^{1/2}} \leq \frac{1}{x^3}$$

But $$\int_2^{\infty} \frac{1}{x^3} dx$$ converges. The Comparison Criterion therefore guarantees that our second integral also converges.

In conclusion, the improper integral converges.

Another input and hope it might be helpful:

Let $$x = \sec(\theta)$$ then $$dx = \tan\theta \sec\theta d\theta$$. With this substitution, we have that $$\sqrt{x^2-1} = \sqrt{\tan^2 \theta}$$ and when $$x = 0 \rightarrow \theta = \sec^{-1}(1) = 0$$, and when $$x \to \infty \rightarrow \theta = \lim_{t \to \infty}\sec^{-1}(t) = \frac{\pi}{2}$$.

Therefore, $$\int_1^{\infty} \frac{1}{x^3\sqrt{x^2-1}} dx = \int_0^{\pi/2} \frac{\sin \theta \cos \theta}{\tan\theta } d\theta = \frac{1}{2} \int_0^{\pi/2} \frac{sin2\theta}{\tan \theta} d\theta \\ = \frac{1}{2} \int_0^{\pi/2} 2\cos^2(\theta) d\theta = \int_0^{\pi/2} \frac{cos 2\theta + 1}{2} d\theta = \frac{\pi}{4}$$

Thus, the improper integral converges.

There is no real problem around $$x=1$$.

Using Taylor expansion $$\frac{1}{x^3\sqrt{x^2 -1}}=\frac{1}{\sqrt{2} \sqrt{x-1}}-\frac{13 \sqrt{x-1}}{4 \sqrt{2}}+O\left((x-1)^{3/2}\right)$$ $$\int\frac{dx}{x^3\sqrt{x^2 -1}}=\sqrt{2} \sqrt{x-1}-\frac{13 (x-1)^{3/2}}{6 \sqrt{2}}+O\left((x-1)^{5/2}\right)$$