Question

Help please!

Answer 1

Question we need. Help you we can.

Answer 2

\[\int\limits_{0}^{\inf}xe^{-x^3} dx \]

Answer 3

sorry im slow with this equation writer

Answer 4

need to transform the integral into
\[\Gamma(p)=\int\limits_{0}^{\inf}x^{p-1} e^{-x} dx, P>0\]

Answer 5

ohhhhhh carappy, I'll need a minute on this one

Answer 6

so, if we let u=(-x^3) du= -3x^2

Answer 7

its fine. these are just tricky substitutions

Answer 8

I think that may work

Answer 9

yeah but you cant get du

Answer 10

I'm not very good with the gamma function

Answer 11

but we can sub it in and get the form we need

Answer 12

I think

Answer 13

I think I tried but it didn't work since you need the x to remain in the function to satisfy gamma function

Answer 14

well, it's sort of like having a dummy variable if I recall correctly.

Answer 15

ill follow your lead :)

Answer 16

So here is what I'm thinking: \[\int\limits_{0}^{\infty}xe^{-x^3} dx=\int\limits_{0}^{\infty}\frac{-1}{3}x^{-1}e^{u} du\]

Answer 17

then we do something like let t=u dt=1

Answer 18

oh crap I forgot to solve for x before, gotta do that first

Answer 19

\[u=-x^3\\ x=-u^{1/3}\]

Answer 20

right, but what can be done about du

Answer 21

so it becomes \[\int\limits_{0}^{\infty}xe^{-x^3} dx=\int\limits_{0}^{\infty}\frac{-1}{3}(-u^{1/3})^{-1}e^{u} du\]

Answer 22

what did you use for du

Answer 23

same as above, I used the x one first simplified then subbed in x in terms of u

Answer 24

hm im not sure i see how you subbed in du; where du= 1/3u^(-2/3)

Answer 25

er sorry dx= 1/3u^(-2/3) du

Answer 26

so, u=-x^3 u'=-3x^2 yea?

Answer 27

yep

Answer 28

so then I subbed it in as is \(dx=\frac{du}{-3x^2}\)

Answer 29

so since I had a x in the thing already, I simplified, then changed my x to a u using \(x=-u^{1/3}\)

Answer 30

oh I see. Okay so then its ready to be transformed

Answer 31

yea, I think so

Answer 32

yeah its in the proper form

Answer 33

so we are here \[\int\limits_{0}^{\infty}xe^{-x^3} dx=\int\limits_{0}^{\infty}\frac{1}{3}(u^{-1/3})e^{u} du\]

Answer 34

I think that should work

Answer 35

the u on e should be negative but other than that this is the proper form

Answer 36

thanks a lot!

Answer 37

well, then modify the sub.

Answer 38

instead of doing u=-x^3 do u=x^3, then it should work nicely

Answer 39

No problem! Happy to help!

Answer 40

okay I will

Answer 41

Have fun!

Answer 42

Thanks :D