integrate: 0 to infinity ʃ e^tx 1/6(e^-x/2 +xe^-x/2)

Question

zarkon · Answer

Finding a Moment Generating Function?

distribute and integrate term by term.

zarkon · Answer

I get $$\frac{3-2t}{3(2t-1)^2}$$

for $$t<1/2$$

anonymous · Answer

yes mgf

anonymous · Answer

how did u know?

zarkon · Answer

I've done a lot of these...could tell by just looking at it.

anonymous · Answer

why is it t<1/2 i dont undertsand

zarkon · Answer

$$\int\limits_{0}^{\infty}e^{tx}e^{-x/2}dx$$

$$\int\limits_{0}^{\infty}e^{tx-x/2}dx$$

$$\int\limits_{0}^{\infty}e^{(t-1/2)x}dx$$

this integral will only converge when (t-1/2)<0  ie t<1/2

anonymous · Answer

so did u take 1/6 out first, then expamded the bracckets and multiplied e^tx with the two exponetials insdie the brackets. is it possible to show me the steps pleeez? :)

zarkon · Answer

yes...pull the 1/6 outside the integral...then distribute e^{tx}...then integrate it term by term...you can do it.

anonymous · Answer

ok i will post my answer..thanks

anonymous · Answer

do u have ti integrate xe^(t-1/2)x by parts?

zarkon · Answer

yes

anonymous · Answer

thnx

zarkon · Answer

no problem