= Solution
Regard $\lambda$ as a <running coupling> $\lambda(M)$. At fixed $s,t,u$ and to order $\lambda^2$, differentiating the one-loop answer with respect to $\log M$ gives
$$
0=\frac{d\Gamma_4}{d\log M}
=\frac{d\lambda}{d\log M}
-\frac{\lambda^2}{32\pi^2}\int_0^1 6\,dx+O(\lambda^3),
$$
because $m\ll M$ implies $F(M^2)\simeq-x(1-x)M^2$. Thus the <beta function (physics)> is
$$
\beta(\lambda)=\frac{d\lambda}{d\log M}
=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).
$$
Solving this <ordinary differential equation> with $\lambda(M_0)=\lambda_0$ gives
$$
\frac1{\lambda(M)}=\frac1{\lambda_0}
-\frac3{16\pi^2}\log\frac{M}{M_0},
\qquad
\lambda(M)=\frac{\lambda_0}{1-\dfrac{3\lambda_0}{16\pi^2}\log(M/M_0)}
$$
to leading-logarithmic order.
Solved by gpt-5.6-sol high.
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