Let $f:\mathbb R_+\to\mathbb R_+$ be defined as $$f(x)=\frac 1{x^n}\text{u}(x)$$ where $n$ is a positive integer and $\text{u}$ is the unit step function. Define $g$ as the convolution of $f$ with itself, i.e. $$g(x)=f(x)*f(x)=\int_0^x f(t)f(x-t)dt$$ Then we have $$\begin{align} g(x)=&\int_0^x t^{-n}(x-t)^{-n}dt\\ \stackrel{\eqref{1}}=&\left(\frac 2x\right)^{2n-1}\int_{-1}^1\left(1-u^2\right)^{-n}du\\ \stackrel{\eqref{2}}=&\left(\frac 2x\right)^{2n-1}\int_0^\pi(\sin\theta)^{1-2n}d\theta\\ \,\\ u&=\frac 2x t-1\tag{1}\label{1}\\ \cos\theta&=u\tag{2}\label{2} \end{align}$$ But $$\begin{align} 0<\theta_0<\pi\implies\sin\theta_0<\theta_0 &\implies\frac 1{\sin \theta_0}>\frac 1\theta_0\\ &\implies\int_0^{\theta_0}\frac {dt}{\sin t}>\int_0^{\theta_0}\frac{dt}t \end{align}$$ Which means the integral is divergent for all positive $n$ and $g(x)$ is undefined. Now the interesting part is, I have tried to evaluate the convolution using Mathematica. For $n=1$ Mathematica gives: $$\frac{\text{u}(x)}x*\frac{\text{u}(x)}x=\frac{\text{u}(x)}x 2\ln{x}$$ and in general, it results in $$g(x)=\frac{b_n+c_n\ln{x}}{x^{2n-1}}\text{u}(x)\qquad n=1,2,3,...$$ where $b_n\le 0$ and $c_n>0$ are constant values. I have already discussed the issue in here. So why am I repeating it here? Good question...

I can't figure out where the mistake is happening. I mean, this wrong result must have been caused by a false assumption somewhere. But I wasn't able to come up with any process that produces such results. The $g(x)$ of Mathematica doesn't make any sense to me. Does anybody have any ideas?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.