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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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    $\begingroup$ 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 $$ \frac{1}{\sqrt{1 + x^2}} \geq \frac{1}{1 + |x|}. $$ $\endgroup$ Commented Aug 22, 2012 at 1:39
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    $\begingroup$ @WillJagy Oh okay, since the improper integral of $ \int^{\infty}_{-\infty}\frac{1}{1+|x|}dx $ diverges, so does my original integral? $\endgroup$ Commented Aug 22, 2012 at 1:49
  • $\begingroup$ Yes ${}{}{}{}{}$ $\endgroup$ Commented Aug 22, 2012 at 1:51
  • $\begingroup$ Got it, thanks! $\endgroup$ Commented Aug 22, 2012 at 1:52
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    $\begingroup$ @Envious Page: (+1) for your question. $\endgroup$ Commented Aug 22, 2012 at 11:09

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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  • $\begingroup$ So the integral diverges because $ \sin(\frac{\pi}{2}) = 1 $ which would result in $ \frac{2}{0} $ which diverges? $\endgroup$ Commented Aug 22, 2012 at 1:51
  • $\begingroup$ Assuming everything else you did was correct, yes. Though I think Will's approach in the comments gets you there faster. $\endgroup$ Commented Aug 22, 2012 at 2:20
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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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  • $\begingroup$ Isn't this already on the page? $\endgroup$ Commented Aug 22, 2012 at 8:12
  • $\begingroup$ @did: sorry? I usually like to post solutions without looking at what the others post. $\endgroup$ Commented Aug 22, 2012 at 8:25
  • $\begingroup$ 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? $\endgroup$ Commented Aug 22, 2012 at 8:27
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    $\begingroup$ 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. $\endgroup$ Commented Aug 22, 2012 at 9:03
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    $\begingroup$ 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. $\endgroup$ Commented Aug 22, 2012 at 13:07

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