= Solution
Let $K$ denote the strike, $\tau=T-t$, and suppose $S>0$, $K>0$, $\tau>0$ and $\sigma>0$. Under the <Black-Scholes model> and the <risk-neutral measure>, $\log S_T$ is normal with
$$
m=\log S+(r-\sigma^2/2)\tau,\qquad v=\sigma\sqrt\tau.
$$
Applying <risk-neutral valuation> to the <European call option> payoff gives
$$
C=e^{-r\tau}\mathbb E_{\mathbb Q}(S_T-K)^+
=e^{-r\tau}\int_{\log K}^{\infty}(e^u-K)\frac{e^{-(u-m)^2/(2v^2)}}{v\sqrt{2\pi}}\,du.
$$
Use $\xi=\Phi((u-m)/v)$, whose differential is precisely the normal density times $du$. The threshold is $\xi_K=\Phi((\log K-m)/v)$, so
$$
\boxed{C=\int_0^1g(\xi)d\xi,\qquad
g(\xi)=e^{-r\tau}\left(e^{m+v\Phi^{-1}(\xi)}-K\right)^+.}
$$
Equivalently, rescale the tail interval to obtain $g_{\mathrm{tail}}(\eta)=e^{-r\tau}(1-\xi_K)\{\exp[m+v\Phi^{-1}(\xi_K+(1-\xi_K)\eta)]-K\}$ on $0<\eta<1$.
For independent uniform inputs, the <Monte Carlo estimator> is $\widehat C_N=N^{-1}\sum_{j=1}^Ng(U_j)$. It is unbiased, converges by the <strong law of large numbers>, and has <variance> $\operatorname{Var}(g(U))/N$. The payoff has finite second moment because the lognormal stock has finite moments. Its estimated standard error is the sample <standard deviation> divided by $\sqrt N$.
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