= Schur evaluation at all ones
{c}
{title2=$s_\lambda(1^N)=\prod_{i<j}\dfrac{\lambda_i-\lambda_j+j-i}{j-i}$}
Put $b_i=\lambda_i+N-i$. In the <bialternant formula>, substitute $x_j=e^{ty_j}$ with distinct $y_j$ and compare the first nonzero <determinant> coefficients as $t\to0$. They arise from the distinct degrees $0,\ldots,N-1$ in the exponential expansions. The common factor in the $y_j$ and the factorials cancel, leaving the product of differences of the $b_i$ divided by the differences of $N-i$. This proves the displayed integral specialization without dividing by a vanishing Vandermonde value.
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