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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?

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    Gerry gave the last step for the way you actually proceeded. Note, meanwhile, that $$ x^2 + 1 \leq x^2 + 2 |x| + 1, $$ so that $$ \frac{1}{\sqrt{1 + x^2}} \geq \frac{1}{1 + |x|}. $$2012-08-22
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    @WillJagy Oh okay, since the improper integral of $ \int^{\infty}_{-\infty}\frac{1}{1+|x|}dx $ diverges, so does my original integral?2012-08-22
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    Yes ${}{}{}{}{}$2012-08-22
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    Got it, thanks!2012-08-22
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    @Envious Page: (+1) for your question.2012-08-22

2 Answers 2

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

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    So the integral diverges because $ \sin(\frac{\pi}{2}) = 1 $ which would result in $ \frac{2}{0} $ which diverges?2012-08-22
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    Assuming everything else you did was correct, yes. Though I think Will's approach in the comments gets you there faster.2012-08-22
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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)

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    Isn't this already on the page?2012-08-22
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    @did: sorry? I usually like to post solutions without looking at what the others post.2012-08-22
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    At what others *already posted*, in the case at hand. Really? Why is that? Aren't you afraid of encumbering the site with 100% duplicate answers?2012-08-22
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    @did: I'm not afraid at all, because I look at that point after finishing my proofs.2012-08-22
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    In other words, you write a proof, then you post it, then you check for dups, then... you do nothing even when you did find some dups?2012-08-22
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    @did: my proof is a bit different because it uses the fact that the integrand is even and I avoid the use of absolute value. Moreover, I've come up with a second solution that brings in place "the new" beside that one similar proof to another already posted as a message and not as an answers.2012-08-22
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    So you are now arguing about the specific merits of your answer when compared to the rest of the page. This seems more reasonable than the principled stance you first took. (To be honest, the arguments you put forth to defend said merits are still a little odd--but this is no big deal.)2012-08-22
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    Whatever. It's still considered polite to acknowledge earlier posts (whether answers or comments) that share ideas with yours, even if you make your post before reading the others.2012-08-22
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    @Gerry Myerson: of course. I appreciate that someone else thought like me. That makes me more confident! I only wanted to emphasize that my approach was a bit different, but this doesn't make it better. I appreciate all people ideas and I usually upvote them all.2012-08-22
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    @did: no. I just pointed out that my answer is a bit different and I considered that it's worth to let it there as a solution.2012-08-22
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    @did: since I have 3 solutions I think I may delete one if you really insist on it.2012-08-22
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    Me? I really do not care. I just wanted to understand a posting strategy which I found (and still find, to be honest) odd. A remark to end this exchange: also odd is to invoke parts of your answer posterior to my first comment (here, your Solution II and Solution III) as a justification to the posting of Solution I.2012-08-22
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    @did: someone that may post 10 different solutions or more for a very simple problem like this one, doesn't need to copy other people's work. (in case you have ever thought at this for a second)2012-08-22
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    This is once again (1.) confusing the temporality of the events on this page and (2.) discussing irrelevant points instead of those explicitly raised. For these reasons and for others, I am very tempted to consider this discussion as closed.2012-08-22
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    @did: yeah. We typed a lot. :-)2012-08-22
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    The way to show that you appreciate that someone else thought like you is to mention her by name and to remark on the ways in which your answer is similar to hers. But I don't get Solution 3 - why is the integral under discussion majorized by the other one? And I have no idea what you're getting at in Solution 4.2012-08-22
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    @GerryMyerson: At the solution 3 I used the fact that $\frac{x}{x^2+1}\leq\frac{1}{\sqrt{x^{2}+1}}$ for all $x\geq0$. Then I take the integrals of both sides and easily prove that the left side goes to infinity that means that the right side tends to infinity, too. I'm not sure if this clarifies your question regarding the solution 3.2012-08-22
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    My mistake - I saw a square root that wasn't there.2012-08-22
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    @Gerry Myerson: I've just improved my first solution.2012-08-22
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    @Gerry Myerson: for the solution 4 I took into account $\int_{0}^{1} \frac{\sqrt{1-x^2}}{x}$ that obviously diverges.2012-08-22
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    @Gerry Myerson: [See here](http://www.wolframalpha.com/input/?i=inverse%281%2Fsqrt%28x^2%2B1%29%29)2012-08-22