Solving Improper Integral of (1/(sqrt(1+x^2))dx

[KevinL]

Homework Statement

Integral from negative infinity to positive infinity of (1/(sqrt(1+x^2))dx

2. The attempt at a solution
Using trig substitution I got the integral equal to ln|sqrt(1+x^2) + x| Finding this was not the difficult part. Evaluating it is.

I set it up like this: lim b --> infinity and lim a --> neg infinity [(ln(sqrt(1+b^2)) + b) - (ln(sqrt(1+a^2)) + a)]

the 'b portion' goes to infinity. For the 'a portion' I rewrote it as ln|1/sqrt(1+x^2) - a| Plug in negative infinity and it is ln|1/infinity|. This is where I am not sure what it is. If 1/infinity = 0, then isn't it indeterminate because you cannot take the ln(0)? If it is simply the ln(extremely small number) then it would be negative infinity, which means the overall answer is infinity, correct?

 

[djeitnstine]

From my analysis the second part seems to be ln(0). Without rewriting it Lim_{x \rightarrow -\infty} ln(\sqrt{1+x^{2}}+x) boils down to Lim_{x \rightarrow -\infty} ln(|x|+x) so a positive plus its negative is always 0.

 

[djeitnstine]

Also I forgot to mention, rewriting it like that would give division by 0

 

[lanedance]

do youknow about hyperbolic trig functions? could be useful here
woops... see you've probably already used them...

 

[lanedance]

you could also look at the symmetry of the integral to re-write it as 2 times the integral from 0 to inf, though this is essentially ln(inf) so infinite itself

but yeah in the limit, the integral tends toward +infinity, as x gets big, the integrand looks like 1/x which is known not to converge

 

[Dick]

Like lanedance said, it doesn't converge. Use a comparison test to show it. Did you misstate the problem?

 

[KevinL]

I definitely stated the problem correctly. In my class we have only gone through all the techniques for integration (u sub, parts, partial fractions, trig sub, etc). We have not done convergence tests or hyperbolic functions.

I see that you can take twice the integral from 0 to infinity since its symmetrical, I just don't understand how you do the limit as a goes to neg infinity of ln(sqrt(1+a^2)) + a)

 

[Dick]

That limit doesn't exist. Neither does the original integral. The area under the curve is infinite. Or lim a->-infinity=-infinity and lim a->+infinity=+infinity. Same thing.