We present various properties of the Erlang loss function B(x,a)=(a∫0∞e-at(1+t)xdt)-1(x≥0,a>0).\documentclass[12pt]{minimal}
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\begin{document}$$B(x,a)=\Bigl( a \int_0^\infty e^{-at} (1+t)^x \,dt \Bigr) ^{-1} \quad{(x\geq 0,\ a>0)}.$$\end{document}
Among other results, we prove:The function
x↦B(x,a)λ\documentclass[12pt]{minimal}
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\begin{document}$$x\mapsto B(x,a)^{\lambda}$$\end{document} is convex on [0,∞)\documentclass[12pt]{minimal}
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\begin{document}$$[0,\infty)$$\end{document} for every a>0\documentclass[12pt]{minimal}
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\begin{document}$$a>0$$\end{document} if and only if λ≤0\documentclass[12pt]{minimal}
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\begin{document}$$\lambda\leq 0$$\end{document} or λ≥1\documentclass[12pt]{minimal}
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\begin{document}$$\lambda \geq 1$$\end{document}.The function x↦(1-B(1/x,a))-1\documentclass[12pt]{minimal}
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\begin{document}$$x\mapsto ({1-B(1/x,a)})^{-1}
$$\end{document} is strictly convex on (0,∞)\documentclass[12pt]{minimal}
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\begin{document}$$(0,\infty)$$\end{document}. This leads to the functional inequality
21-B(H(x,y),a)<11-B(x,a)+11-B(y,a)(x,y>0,x≠y),\documentclass[12pt]{minimal}
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\begin{document}$$\frac{2}{1-B( H(x,y) ,a)}<
\frac{1}{1-B(x,a)} +\frac{1}{1-B(y,a)}\quad{(x,y>0,\ x\neq y)},
$$\end{document}
where H(x,y)=2xy/(x+y)\documentclass[12pt]{minimal}
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\begin{document}$$H(x,y)=2xy/(x+y)$$\end{document} denotes the harmonic mean of x and y.
Let a>0\documentclass[12pt]{minimal}
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\begin{document}$$a>0$$\end{document}. The inequality B(x,a)+B(1/x,a)≤1\documentclass[12pt]{minimal}
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\begin{document}$$B(x,a) + B(1/x,a)\leq 1$$\end{document} holds for all x>0\documentclass[12pt]{minimal}
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\begin{document}$$x>0$$\end{document} if and only if a≤1\documentclass[12pt]{minimal}
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\begin{document}$$a\leq 1$$\end{document}.
