A doubt in integration/improper integration theory

[dunkelheit]

While I was evaluating

[MATH]\int_{-1}^{1} \frac{\cos x}{e^{\frac{1}{x}}+1} \text{d}x[/MATH]​

I had a doubt: the function [MATH]f(x):=\frac{\cos x}{e^{\frac{1}{x}}+1}[/MATH] is defined in [MATH][-1,0) \cup (0,1][/MATH], but [MATH]x=0[/MATH] is in the integration interval; so to eliminate this problem I can define a new function

[MATH]g(x):= \begin{cases} f(x), & \text{if} \ x \in[-1,1] \setminus \{0\} \\ 0, & \text{if} \ x=0 \end{cases}[/MATH]​

So, since the value of [MATH]g[/MATH] at [MATH]x=0[/MATH] doesn't change the value of the integral (since it is a single point), we have that

[MATH]\int_{-1}^{1} \frac{\cos x}{e^{\frac{1}{x}}+1} \text{d}x=\int_{-1}^{1} g(x) \text{d}x[/MATH]​

However I could do this argument with functions like [MATH]\frac{1}{x}[/MATH], so I'm not sure if what I've said is correct. Maybe the fact that [MATH]\frac{1}{x}[/MATH] is not bounded in a neighbourhood of [MATH]x=0[/MATH] means something? It has to do with improper integrals?
So the question is: when can I do the first argument (if it is correct) and define a new function such that the integral doesn't change and when it is not allowed?
Thanks.

 

[Steven G]

Use limits. If this hint is not helpful then please ask for another hint.

 

[LCKurtz]

It's easy enough to see the integral converges since [math]\left | \frac {\cos x}{e^{\frac 1 x} + 1}\right | \le 1[/MATH] on that interval. Are you expecting to actually evaluate it? Good luck with that...

 

[dunkelheit]

@Jomo: I don't get your hint, sorry. Thanks for your answer.
@LCKurtz: Thanks for your answer. The evaluation is not that hard actually, letting [MATH]x=-y[/MATH] you can see that

[MATH]I=\int_{-1}^{1} \frac{\cos x}{e^{\frac{1}{x}}+1} \text{d}x=\int_{-1}^{1} \frac{\cos y}{e^{-\frac{1}{y}}+1} \text{d}y=\int_{-1}^{1} \frac{e^{\frac{1}{y}} \cos y }{e^{\frac{1}{y}}+1} \text{d}y \Rightarrow 2I=\int_{-1}^{1} \cos x \text{d}x=2 \sin 1 \Rightarrow I=\sin 1[/MATH]​

By doubt was in integration theory, why this integral can be evaluated in this way even if the integrand is not defined in [MATH]x=0[/MATH] and in other cases, like [MATH]\frac{1}{x}[/MATH], it can't be done.

 

[topsquark]

I didn't try to derive it, but I have little doubt that what you are reporting is not the value of the integral, but the "principle value" of the integral.
Say that the f(x) has some kind of discontinuity at x = c, where b < c < a. Then [math]\int_{b}^a f(x) ~ dx \to P\int_{b}^a f(x) ~ dx = \lim_{e \to c} \int_b^e f(x) ~ dx + \lim_{e \to c} \int_e^a f(x) ~ dx[/math]. Essentially we cut the pole out of the region of integration.

For example
[math]P\int_{-1}^2 \dfrac{1}{x} ~ dx = \lim_{a \to 0} \int_{-1}^a \dfrac{1}{x} ~ dx + \lim_{a \to 0} \int_a^2 \dfrac{1}{x} ~ dx[/math]
[math]= \lim_{a \to 0} ( ln|a| - ln|-1| ) + \lim_{a \to 0} ( ln|2| - ln|a| )[/math]
The ln|a| terms cancel out leaving [math]-ln(1) + ln(2) = ln(2)[/math].

-Dan

 

[dunkelheit]

@topsquark: Thanks, but actually the function [MATH]\frac{\cos x}{e^{\frac{1}{x}}+1}[/MATH] is not discontinuous at [MATH]x=0[/MATH] because it isn't defined for [MATH]x=0[/MATH]; it is continuous in all [MATH][-1,0) \cup (0,1][/MATH]. Same for [MATH]\frac{1}{x}[/MATH], it is continuous in all [MATH](-\infty,0) \cup (0,\infty)[/MATH] and so it is continuous in all its domain, but it is unbounded in a neighbourhood of [MATH]x=0[/MATH] and this makes its integral improper, things that doesn't happen with [MATH]\frac{\cos x}{e^{\frac{1}{x}}+1}[/MATH] since it is bounded in a neighbourhood of [MATH]x=0[/MATH].

 

[topsquark]

@topsquark: Thanks, but actually the function [MATH]\frac{\cos x}{e^{\frac{1}{x}}+1}[/MATH] is not discontinuous at [MATH]x=0[/MATH] because it isn't defined for [MATH]x=0[/MATH]; it is continuous in all [MATH][-1,0) \cup (0,1][/MATH]. Same for [MATH]\frac{1}{x}[/MATH], it is continuous in all [MATH](-\infty,0) \cup (0,\infty)[/MATH] and so it is continuous in all its domain, but it is unbounded in a neighbourhood of [MATH]x=0[/MATH] and this makes its integral improper, things that doesn't happen with [MATH]\frac{\cos x}{e^{\frac{1}{x}}+1}[/MATH] since it is bounded in a neighbourhood of [MATH]x=0[/MATH].

If the result is due to taking the principal value it doesn't matter as we are taking limits and the point x = 0 never appears in the integral. If not then, yes, there is a problem. It's my best guess as to what is going on.

-Dan

 

[Steven G]

Since the integrand is not defined at x=0 it is customary to change the integral into a sum of two integrals:

Integral from x=-1 to k + integral from x=k to x=1 and then take the appropriate limits.

 

[LCKurtz]

@LCKurtz: Thanks for your answer. The evaluation is not that hard actually, letting [MATH]x=-y[/MATH] you can see that

[MATH]I=\int_{-1}^{1} \frac{\cos x}{e^{\frac{1}{x}}+1} \text{d}x=\int_{-1}^{1} \frac{\cos y}{e^{-\frac{1}{y}}+1} \text{d}y=\int_{-1}^{1} \frac{e^{\frac{1}{y}} \cos y }{e^{\frac{1}{y}}+1} \text{d}y \Rightarrow 2I=\int_{-1}^{1} \cos x \text{d}x=2 \sin 1 \Rightarrow I=\sin 1[/MATH]​

That's a pretty clever argument to evaluate it. I don't think I have seen that before, at least not in this setting.

By doubt was in integration theory, why this integral can be evaluated in this way even if the integrand is not defined in [MATH]x=0[/MATH] and in other cases, like [MATH]\frac{1}{x}[/MATH], it can't be done.

It has been many years since I looked in a real analysis book so I don't recall the exact general conditions to answer your question. But for this particular example, the function is obviously well behaved at the origin, having both right hand limits for the function and its derivative at [MATH]0[/MATH]. Pretty clearly Riemann integrable regardless of whether or not it has a value at the jump discontinuity.