@JohnDCook generally:
int_{0}^{1} u(x) dx = int_{0}^{\inf} u(ξ) dξ
x = exp(-ξ), dx= - exp(-ξ) dξ
I call this a 'procrastination'
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@JohnDCook generally:
int_{0}^{1} u(x) dx = int_{0}^{\inf} u(ξ) dξ
x = exp(-ξ), dx= - exp(-ξ) dξ
I call this a 'procrastination'