Improper integral of an even function

In summary, the colleagues discussed the integration of even functions, specifically when the function is not Riemann integrable but is Cauchy integrable. They explored the example of the function ${x}^{2}\ln\left| x \right|$ and concluded that it can be integrated using a limit and can be written as $\int_{-1}^1 x^2\ln(|x|)dx=2\lim_{\epsilon\to 0^+}\int_\epsilon^1 x^2 \ln(|x|)dx= 2\int_0^1 x^2 \ln(|x|) dx$, as long as the limit exists.
  • #1
sarrah1
66
0
Hi colleagues

This is a very very simple question

I can show when $f$ is integrable and is even i.e. $f(-x)=f(x)$ then

$\int_{-a}^{a} \,f(x)\,dx=2\int_{0}^{a} \,f(x)\,dx$

what about improper integrals of even functions, like the function ${x}^{2}\ln\left| x \right|$ this function is even but undefined at the origin. Yet it has a limit equal to zero there. I wish to integrate it from $[-1,1]$. I know that I can partition the integral into $[-1,0]$ then $[0,1]$ and in each interval I can carry on the improper integration. My simple question is since the function is not Riemann integrable, but Cauchy integrable can I still write

$\int_{-1}^{1} \,{x}^{2}\ln\left| x \right|\,dx=2\int_{0}^{1} \,{x}^{2}\ln\left| x \right|\,dx=2\lim_{{c}\to{{0}^{+}}}\int_{c}^{1} \,{x}^{2}\ln\left| x \right|\,dx$

Many thanks
Sarrah
 
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  • #2
You know, I presume, that an "improper integral" is a limit. Here, we have $\int_{-1}^1 x^2\ln(|x|)dx= \lim_{\epsilon\to 0^-} \int_{-1}^\epsilon x^2\ln(|x|)dx+ \lim_{\epsilon\to 0^+}\int_{\epsilon}^1 x^2 \ln(|x|)dx$.

Since those two integrals avoid the singularity at x= 0, yes, we can write it as
$\int_{-1}^1 x^2\ln(|x|)dx=2\lim_{\epsilon\to 0^+}\int_\epsilon^1 x^2 \ln(|x|)dx= 2\int_0^1 x^2 \ln(|x|) dx$

(Provided, of course, the limit exists.)
 
  • #3
HallsofIvy said:
You know, I presume, that an "improper integral" is a limit. Here, we have $\int_{-1}^1 x^2\ln(|x|)dx= \lim_{\epsilon\to 0^-} \int_{-1}^\epsilon x^2\ln(|x|)dx+ \lim_{\epsilon\to 0^+}\int_{\epsilon}^1 x^2 \ln(|x|)dx$.

Since those two integrals avoid the singularity at x= 0, yes, we can write it as
$\int_{-1}^1 x^2\ln(|x|)dx=2\lim_{\epsilon\to 0^+}\int_\epsilon^1 x^2 \ln(|x|)dx= 2\int_0^1 x^2 \ln(|x|) dx$

(Provided, of course, the limit exists.)

thank you very much
 

Related to Improper integral of an even function

1. What is an improper integral of an even function?

An improper integral of an even function is an integral where the limits of integration are infinite or the integrand is unbounded at one or both of the limits. It is called an even function because the function being integrated is symmetrical about the y-axis.

2. How is an improper integral of an even function calculated?

An improper integral of an even function can be calculated by using the properties of even functions, such as symmetry, to simplify the integral. This can lead to the integral being rewritten in terms of a definite integral, which can then be evaluated using standard integration techniques.

3. What is the significance of an even function in an improper integral?

An even function is significant in an improper integral because it allows for simplification and easier evaluation of the integral. Additionally, the symmetry of the function ensures that the integral will converge, meaning that the area under the curve will have a finite value.

4. Can an improper integral of an even function be divergent?

No, an improper integral of an even function cannot be divergent. This is because the symmetry of the function ensures that the integral will converge, even if the limits of integration are infinite or the integrand is unbounded.

5. What are some real-world applications of an improper integral of an even function?

An improper integral of an even function has many applications in physics, engineering, and economics. It can be used to calculate the total energy of a system, the area under a symmetrical probability distribution, or the average value of a periodic function. It is also commonly used in Fourier analysis and in solving differential equations.

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