L'Hôpital's Rule for $ \frac{\infty}{\infty-\infty} $?

Question

I'm currently trying to find this improper integral: $$ \int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx $$

I started off by splitting it into a proper integral, and then into the sum of two integrals: $$ = \lim_{a\rightarrow\infty} \int^{a}_{-a}\frac{1}{\sqrt{x^{2}+1}}dx = \lim_{a\rightarrow\infty}(\int^{0}_{-a}\frac{1}{\sqrt{x^{2}+1}}dx + \int^{a}_{0}\frac{1}{\sqrt{x^{2}+1}}dx) $$

To calculate the integrals I used the trig. substitution $ x=b\tan\theta $ with $ b=1 $, which would give the differential $ dx=sec^{2}\theta d\theta $. The new limits of integration would then be $ [-\frac{\pi}{2},0] $ and $ [0,\frac{\pi}{2}] $ because as $ x\rightarrow\pm\infty $, $ \theta\rightarrow\pm\frac{\pi}{2} $, so the integrals and limit can be rewritten as: $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sqrt{\tan^{2}\theta+1}}d\theta + \int^{a}_{0}\frac{\sec^{2}\theta}{\sqrt{\tan^{2}\theta+1}}d\theta) $$

...which can then simplify to: $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sqrt{\sec^{2}\theta}}d\theta +\int^{a}_{0}\frac{\sec^{2}\theta}{\sqrt{\sec^{2}\theta}}d\theta) = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{|\sec\theta|}d\theta+\int^{a}_{0}\frac{\sec^{2}\theta}{|\sec\theta|}d\theta) $$

The absolute values on the secants can be removed because on the interval $ [-\frac{\pi}{2},\frac{\pi}{2}] $, the secant function is positive. $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sec\theta}d\theta+\int^{a}_{0}\frac{\sec^{2}\theta}{\sec\theta}d\theta) = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\sec\theta d\theta+\int^{a}_{0}\sec\theta d\theta) $$

The antiderivative of $ \sec\theta = \ln|\sec\theta+\tan\theta|+C $, so the integrals become: $$ = \lim_{a\rightarrow\pi/2}(\ln|\sec\theta+\tan\theta|\bigg|^{0}_{-a} + \ln|\sec\theta+\tan\theta|\bigg|^{a}_{0}) $$ $$ = \lim_{a\rightarrow\pi/2}((\ln|\sec(0)+\tan(0)|-\ln|\sec(-a)+\tan(-a)|)+(\ln|\sec(a)+tan(a)|-\ln|\sec(0)+tan(0)|)) $$

Since $ \sec(0) = 1 $ and $ \tan(0) = 0 $, the value of $ \ln|\sec(0)+tan(0)| = \ln(1) = 0 $. The limit can be rewritten as: $$ = \lim_{a\rightarrow\pi/2}((0-\ln|\sec(-a)+\tan(-a)|)+(\ln|\sec(a)+tan(a)|-0)) $$ $$ = \lim_{a\rightarrow\pi/2}(-\ln|\sec(-a)+\tan(-a)|+\ln|\sec(a)+tan(a)|) $$

The tangent function has been shown to be odd, and the secant function even, so $ \sec(-a) = \sec(a) $ and $ \tan(-a) = -\tan(a) $. Therefore, applying and then commuting the addition, we have: $$ = \lim_{a\rightarrow\pi/2}(\ln|\sec(a)+tan(a)|-\ln|\sec(a)-\tan(a)|) $$

Subtraction of logarithms become division, so $ \ln|\sec(a)+tan(a)|-\ln|\sec(a)-\tan(a)| $ $ = \ln\left|\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right| $, which becomes: $$ = \lim_{a\rightarrow\pi/2}\left(\ln\left|\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right|\right)$$

Here's where I'm confused: can you take the natural log of the limit of the fraction (i.e., $$ \ln\left|\lim_{a\rightarrow\pi/2}\left(\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right)\right| $$ ), or does the limit not exist? And, if you can take the natural log of the limit, how would you go about evaluating the limit of the fraction? Since $ \sec(\frac{\pi}{2}) "=" \infty $ and $ \tan(\frac{\pi}{2}) "=" \infty $, would there be some form of L'Hôpital's Rule you'd have to use, since $ \frac{\infty}{\infty-\infty} $ is indeterminate?

Answer 1

$${\sec x+\tan x\over\sec x-\tan x}={1+\sin x\over1-\sin x}$$

Answer 2

Solution I

Note that the integrand is even and then you have that: $$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ but $$\int^{\infty}_{0}\frac{1}{x+1} \longrightarrow \infty\leq\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ and the improper integral diverges.

This first solution is very similar to Will Jagy's solution you may find in a message above.

Q.E.D.

Solution II

Also observe that the integrand is the derivative of $\sinh^{-1}$(x). The conclusion is evident.

Q.E.D.

Solution III

Another elementary solution?

$$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ $$\int^{\infty}_{0}\frac{x}{x^2+1}= \lim_{x\to\infty}\frac{1}{2} \ln (x^2+1) \longrightarrow \infty\leq\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$

Q.E.D.

Solution IV

Could the inverse of the integrand allow us to evaluate the improper integral without being necessary to use any integration? (see the real positive axes)

Solution V

Consider again that $$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$

then you do 2 things. Firstly, note $x = \tan y$ and for the result you get, use the nice work of Raymond Manzoni here, namely the first 3 rows of his answer and you're nicely done.

(of course, it is enough to compute the limit to $\frac{\pi}{2}$, but the approach from the link is worth to be seen)