Any idea on how to integrate the following?

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SUMMARY

The discussion focuses on finding the moment generating function (M(t)) for a random variable X with a cumulative distribution function (c.d.f.) defined as F(x) = 1 - e^(-x²) for x ≥ 0. The probability density function (pdf) is derived as f(x) = F'(x) = 2xe^(-x²). To compute M(t), the integral M(t) = E(e^(tX)) = ∫ e^(tx) 2xe^(-x²) dx is proposed, with a suggestion to complete the square in the exponent to simplify the integration process.

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Homework Statement



The c.d.f. of a random variable X is given by the following formula

F(x) = 1 - e-x2 x≥0

Find M(t). Moment generating function.

The Attempt at a Solution



I've found the pdf as

f(x) = F'(x) = 2xe-x2 x≥0

To determine M(t) I need to do the following

M(t) = E(etX) = ∫etx 2xe-x2

Then I'm not sure how to go about integrating. Any ideas or clues to point me in the right direction will be greatly appreciated. Thanks.
 
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Try completing the square: ##x^2 - tx = x^2 - tx + (t/2)^2 - (t/2)^2 = (x - t/2)^2 - (t/2)^2##, so
$$\int e^{tx}2xe^{-x^2}dx = e^{(t/2)^2} \int 2x e^{-(x-t/2)^2} dx$$
which should be manageable with a bit more work.
 
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