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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?

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