= Solution
Apply the <Dimensional regularization>[dimensionally regulated] <Tadpole diagram>[tadpole] integral before removing the <Infrared regularization>[infrared regulator]:
$$
I_\gamma(m_\gamma)=\mu^\epsilon\int\frac{d^dl}{(2\pi)^d}\frac1{l^2+m_\gamma^2}
=\frac{i\mu^\epsilon}{(4\pi)^{2-\epsilon/2}}\Gamma\left(-1+\frac\epsilon2\right)(m_\gamma^2)^{1-\epsilon/2}.
$$
At fixed regulator $0<\operatorname{Re}\epsilon<2$ away from <Gamma function>[gamma-function poles], the power of $m_\gamma^2$ has positive real part. Consequently
$$
\boxed{\lim_{m_\gamma\to0}I_\gamma(m_\gamma)=0.}
$$
Both the correctly weighted <Tadpole diagram>[tadpole] and the source's doubled version vanish. This is the near-four-dimensional limit appropriate to the question; the order of limits is important. Equivalently the exactly massless <Tadpole diagram>[tadpole] is a <scaleless integral> and is zero in <dimensional regularization>. The statement is not valid for every arbitrary fixed dimension: for example, $\epsilon=5/2$ gives a nonzero prefactor and growth as $m_\gamma^{-1/2}$. One must keep the regulator in its stated near-four-dimensional domain.
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