Improper Integral ... Help!

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trojan

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Feb 7, 2007
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I'm a tutor and I ran into a student's homework problem that I can't crack. The question is:

Integral from negative infinity to positive infinity of the funtion 1 / sqrt [x^4 + 1]

Does this converge or diverge? Can the integral be evaluated? This section was on improper integrals. At first I wanted to compare that Integral to 1 / x^2 as it is always bigger, but then that function isn't continuous at 0, so I don't think I can do that. Then I wanted to see if it would diverge and so I compared it to 1 / (x^2 +1) as that function is always smaller, but I believe that evaluates out to arctan x and that converges to pi, so I can't say the original function diverges. I can't see any "u" substitutions and am now quite frustrated. Any help would be appreciated.
 
The integral \(\displaystyle \int\limits_{ - \infty }^\infty {\frac{{dx}}{{x^2 + 1}}}\)
converges and has value \(\displaystyle \pi\).
 
Thank you for your quick response, but could you step by step illustrate how to solve / prove it converges to pi? The steps in solving is what I need.
 
I don't know if this is a good idea. Pka may scold me, but:

\(\displaystyle \L\\\int\frac{1}{x^{2}+1}dx=tan^{-1}(x)\)

\(\displaystyle \L\\tan^{-1}(\infty)=\frac{\pi}{2}\)
 
Consider this \(\displaystyle x \ge 1 \Rightarrow \quad \int\limits_1^B {\frac{{dx}}{{\sqrt {x^4 + 1} }}} \le \int\limits_1^B {\frac{{dx}}{{x^2 }}} = 1 - \frac{1}{B}\) and \(\displaystyle 0 < x < 1 \Rightarrow \quad \sqrt {x^4 + 1} > \sqrt x \Rightarrow \quad \frac{1}{{\sqrt {x^4 + 1} }} < \frac{1}{{\sqrt x }}\).

These means that \(\displaystyle \int\limits_0^\infty {\frac{{dx}}{{\sqrt {x^4 + 1} }}}\) exists. Note you have an even function.

I cannot give you the exact value, but it does converge.
 
Hello. Perhaps I'm not asking my question properly. I want to integrate 1 / sqrt [x^4 +1], not 1 / [x^2 +1]. Those two equations are not the same.

thanks,
 
I understand that! Look at my second post.
 
If I may, \(\displaystyle \L\\\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^{4}+1}}dx=3.081493546\)

I don't believe this is integrable by elementary means. Was the point of the problem to just see if it converges or diverges?. That's what pka showed.
 
I don't quite understand why you needed to compare the orginal function to 1 / x^2 on the interval from 1 to infinity and then compare it to 1 / sqrt x on interval 0 to 1. 1 / x^2 is already always bigger from the whole interval 0 to infinity. 1 / x^2 is actually bigger over negative infinity to positive infinity, so if I could prove 1 / x^2 converges then I could say the original function also converges. But the problem I have is that 1 / x^2 is discontinuous at 0. So how do I get around that?
 
Okay, I think the point now is to determine if converge or diverge, but I'm still not exactly satisfied with the proof of convergence. Maybe I'm missing something. Please see the post I put above this one.
 
Thank you PKA. After reading your posts again and again and again, it finally hit me. I understand why you now broke it up over two intervals 0 to 1 and 1 to infinity and compared the original equation to 1 / sqrt x and 1 / x^2. Doing it that way gets rid of the undefined problem of 1 / x^2 being evaluated at 0.

Thank you guys for your mastery!
 
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