dunkelheit
New member
- Joined
- Sep 7, 2018
- Messages
- 48
While I was evaluating
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.
[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.