- #1
noospace
- 69
- 0
This is not a homework question, just trying to understand the material in my differential geomety lecture.
Consider the function
f : \mathbb{R} \to [0,1]
given by f(x) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt
where a(x) is zero for x less than or equal to 0 and a(x) = e^{-x^{-1}} for x > 0 and A = \int_{0}^{1}a(t)a(1-t)dt.
The claim is that f(x) = 0 for x less than or equal to 0 and f(x) = 1 for x greater than or equal to 1.
The first part is obvious. The second part I'm having trouble with.
f(x > 1) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt
f(x > 1) = \frac{1}{A}\left( \int_{-\infty}^0 + \int_{0}^{1} + \int_{1}^x \right) a(t) a(1-t)dt
f(x > 1) = 1 + \frac{1}{A} \int_{1}^x a(t) a(1-t)dt.
So the question becomes, why does the second term vanish? My lecturer claims this is obvious but I just don't see it.
Consider the function
f : \mathbb{R} \to [0,1]
given by f(x) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt
where a(x) is zero for x less than or equal to 0 and a(x) = e^{-x^{-1}} for x > 0 and A = \int_{0}^{1}a(t)a(1-t)dt.
The claim is that f(x) = 0 for x less than or equal to 0 and f(x) = 1 for x greater than or equal to 1.
The first part is obvious. The second part I'm having trouble with.
f(x > 1) = \frac{1}{A}\int_{-\infty}^{\infty} a(t) a(1-t)dt
f(x > 1) = \frac{1}{A}\left( \int_{-\infty}^0 + \int_{0}^{1} + \int_{1}^x \right) a(t) a(1-t)dt
f(x > 1) = 1 + \frac{1}{A} \int_{1}^x a(t) a(1-t)dt.
So the question becomes, why does the second term vanish? My lecturer claims this is obvious but I just don't see it.