3 ms·
The second equation of Section 1 says one can "easily" see how ∫0->∞ of x*e^-tx dx equals -1/t², but I'm just not seeing it. Can somebody help me out here
by raldi 6y ago
The second equation of Section 1 says one can "easily" see how
∫0->∞ of x*e^-tx dx
equals -1/t², but I'm just not seeing it. Can somebody help me out here?
- mvanaltvorst 6y agoYou differentiate both sides of the equation given above with respect to t. Differentiating 1/t yields -1/t^2. On the LHS you apply the differentiation inside the integral.
- greens 6y ago∫0->∞ x*e^-tx dx = -∂_t ∫0->∞ e^-tx dx = -∂_t (1/t)∫0->∞ e^-u du = -∂_t (1/t) = 1/t^2