Improper Intregral Question

[Jon Blatt]

Hi,
So heres the question

the integral of tan(x) with the bounds being 0 to pi/2. I know this is a improper integral with pi/2 being the infinite asymptote...I get stuck when i take the limit of the antiderivate lim=ln(sec (a)) - ln(sec (0)) where the limit goes from a-->pi/2 from the left ....im stuck here. thanks for any help...

 

[Count Iblis]

Hi,
So heres the question

the integral of tan(x) with the bounds being 0 to pi/2. I know this is a improper integral with pi/2 being the infinite asymptote...I get stuck when i take the limit of the antiderivate lim=ln(sec (a)) - ln(sec (0)) where the limit goes from a-->pi/2 from the left ....im stuck here. thanks for any help...

You just expand cos(x) around x = pi/2:

cos(pi/2 - epsilon) = sin(epsilon)= epsilon -epsilon^3/6 + ... --->

Log[cos(pi/2 - epsilon)] = Log[epsilon -epsilon^3/6 + ... ]=

Log[epsilon] + Log[1-epsilon^2/6 + ...]

Then expand the Logarithm

Log[1 + epsilon] = epsilon - epsilon^2/2 + ... ---->

Log[1-epsilon^2/6 + ...] = -epsilon^2/6+...

We thus have:

Log[cos(pi/2 - epsilon)] = Log[epsilon] --epsilon^2/6 +...

And you see that the integral diverges logarithmically in the limit epsilon --> 0

 

[Jon Blatt]

umm...you lost me...Im not familiar with that technique of finding the limit...I am in Calculus 2 in college right now...is there a better way of explaining it?

 

[galactus]

Occam's Razor:

\(\displaystyle \L\\\int_{0}^{\beta}tan(x)dx=ln|sec(x)|=ln|sec({\beta})|\rightarrow{\infty}, \;\ as \;\ {\beta}\rightarrow\frac{\pi}{2}^{-}\)...Divergent.

 

[Count Iblis]

umm...you lost me...Im not familiar with that technique of finding the limit...I am in Calculus 2 in college right now...is there a better way of explaining it?

You should study Taylor expansion methods first, even if it is taught at a later stage. To calculate limits, you need to know how a function behaves near the limit point, this is given by the series expansion. Now, you can calculate limits using high school methods also, but that's cumbersome in general. To see the connection with series expansions, consider the limits.


Lim x-->0 Sin(x)/x = 1

Lim x -->0 [Sin(x) - x]/x^3 = -1/6

Lim x -->0 [Sin(x) - x + x^3/6]/x^5 = 1/120


The first limit essenstially says that for x near zero sin(x) is approximately x. If we subtract x from sin(x), then that function should go to zero much faster. It turns out that it goes to zero like x^3, so if you divide it by x^3 you get a finite limit. That limit turns out to be -1/6. This means that near zero sin(x) -x is approximately -1/6 x^3. If you subtract that then you get a function thast goes to zero even faster. All the above limits can be calculated using high school methods, e.g. L'Hopital's rule. But I think that you see now that if you know how sin(x) behaves near x=0:


Sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ....

You can calculate all sorts of limits there without much effort.