Improper integral of an even function

[sarrah1]

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

 

[HallsofIvy]

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.)
​

 

[sarrah1]

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