Finite UV subtraction and a divergent factorial toy series

The integrand $r\mapsto r^3\mathrm{/}(r^2+{\mu }^2)$ is nonnegative and continuous on $[0,\mathrm{\Lambda }]$. Apply [F2], then divide by $(2\pi {)}^4$, to obtain $I_{\mathrm{\Lambda }}=\frac{1}{8{\pi }^2}{\int }_0^{\mathrm{\Lambda }}\frac{r^3}{r^2+{\mu }^2}dr\mathrm{.}$ Polynomial division gives $r^3\mathrm{/}(r^2+{\mu }^2)=r-{\mu }^{2}r\mathrm{/}(r^2+{\mu }^2)$. By [F4], an antiderivative is $G(r)=r^2\mathrm{/}2-({\mu }^2\mathrm{/}2)\log (r^2+{\mu }^2)$ on this interval. Its derivative is continuous; its compact-interval integrands are bounded and Riemann integrable, so [F9] identifies their Lebesgue integrals with the calculus integrals. Thus [F3] evaluates the integral as $G(\mathrm{\Lambda })-G(0)=\frac{1}{2}({\mathrm{\Lambda }}^2-{\mu }^2\log (1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2))\mathrm{.}$ Multiplying by $1\mathrm{/}(8{\pi }^2)$ gives the claimed expression.

By [F6], $-\exp (-x)$ has derivative $\exp (-x)$; [F3] on $[0,R]$ gives ${\int }_0^R\exp (-x)dx=1-\exp (-R)$. The positive series [F5] gives $\exp (R)\ge 1+R$, so $\exp (-R)\to 0$ and $M_0=1$. For each integer $n\ge 0$, [F5] with $y=x\mathrm{/}2$ gives $x^n\exp (-x)\le 2^{n}n!\exp (-x\mathrm{/}2)$. The majorant has finite integral $2^{n+1}n!$ by [F3] and [F6]; the positive truncated integrals are increasing and bounded by this finite integral, hence have a finite improper limit. All the integrands are continuous on every compact interval; [F9] identifies these convergent nonnegative improper integrals with their half-line Lebesgue integrals. Thus $M_n={\int }_0^{\mathrm{\infty }}{x}^n\exp (-x)dx$ is finite. The same bound and $\exp (x\mathrm{/}2)\ge 1+x\mathrm{/}2$ show $x^n\exp (-x)\to 0$. For $n\ge 1$ and $R>0$, take $u(x)=x^n$, $v(x)=-\exp (-x)$ on $[0,R]$. They are continuously differentiable with continuous, hence integrable, derivatives. Thus [F7] gives ${\int }_0^R{x}^n\exp (-x)dx=-R^n\exp (-R)+n{\int }_0^R{x}^{n-1}\exp (-x)dx\mathrm{.}$ The endpoint at zero vanishes since $n\ge 1$. Taking $R\to \mathrm{\infty }$ is justified by the finite positive improper integrals and the vanishing boundary term just proved. It gives $M_n=n{M}_{n-1}$, and induction from $M_0=1$ gives $M_n=n!$. Finally $0\le \exp (-x)\mathrm{/}(1+\lambda x)\le \exp (-x)$ for $\lambda \ge 0$, so $F(\lambda )$ is finite.

Substituting the formula from step 1.1, the subtracted quantity equals $\frac{{\mu }^2}{16{\pi }^2}\log \left(\frac{{\mu }^2}{{\mu }_R^2}\frac{{\mathrm{\Lambda }}^2}{{\mathrm{\Lambda }}^2+{\mu }^2}\right)\mathrm{.}$ As $\mathrm{\Lambda }\to \mathrm{\infty }$, the second factor inside the logarithm tends to $1$; the limiting logarithm is defined because ${\mu }^2\mathrm{/}{\mu }_R^2>0$. Continuity of $\log$ on $(0,\mathrm{\infty })$ proves the stated finite limit.

For every $t\ge 0$, finite multiplication of a geometric sum gives $\frac{1}{1+t}={\sum }_{n=0}^N(-t{)}^n+(-1{)}^{N+1}\frac{t^{N+1}}{1+t}\mathrm{.}$ Put $t=\lambda x$, multiply by $\exp (-x)$, and integrate. Step 1.2 evaluates each polynomial moment as $n!$, while the remainder is bounded by ${\lambda }^{N+1}{M}_{N+1}={\lambda }^{N+1}(N+1)!$. This gives the exact finite identity and its stated bound, including $\lambda =0$.

For each fixed $\lambda >0$, the absolute terms $n!{\lambda }^n$ in the formal real power series have successive ratio $(n+1)\lambda$, eventually greater than $2$. They therefore do not tend to zero, so the series does not converge at any positive $\lambda$. By [F10], no $r>0$ has absolute convergence at every $\mathrm{\mid }\lambda \mathrm{\mid }<r$: choosing $\lambda =r\mathrm{/}2$ contradicts the divergence just proved. The only admissible radius is $r=0$, so the supremum in [F10] is zero. Separately, the bare positive coefficient grows quadratically because $\frac{I_{\mathrm{\Lambda }}}{{\mathrm{\Lambda }}^2}=\frac{1}{16{\pi }^2}(1-\frac{{\mu }^2}{{\mathrm{\Lambda }}^2}\log (1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2))⟶\frac{1}{16{\pi }^2}\mathrm{.}$ For the last limit, with $y=1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2$, [F4] gives $\log y={\int }_1^{y}du\mathrm{/}u\le {\int }_1^{y}du\mathrm{/}\sqrt{u}\le 2\sqrt{y}$, so the logarithmic term divided by ${\mathrm{\Lambda }}^2$ tends to zero.

Steps 1.1 and 2.1 give the exact cutoff integral and selected finite subtraction limit; steps 1.2 and 2.2 prove the finite factorial remainder; step 2.3 proves that the formal power series diverges for every $\lambda >0$. These conclusions concern only the two displayed mathematical integrals and do not assert a full QED renormalization theorem or a continuum limit.