Question

Express the following Gamma functions and evaluate them using a table of Gamma function

Answer 1

\[\int\limits_{0}^{1} x^{2} \left( \ln \frac{ 1 }{ x } \right)^{3} dx \]
HINT : Put x = \(e^{-u}\)

Answer 2

Are you able to make the substitution sucessfully?

Answer 3

No.., i'm not..,

Answer 4

Note that
x=e^(-u) means
ln(x)=-u
-ln(x)=u
ln(1/x)=u
Remembe to change your dx as well

Answer 5

ok.., i'll try

Answer 6

i would use integration by parts about 3, or 4 times

Answer 7

actually we might be able to find a reduction formula for this

Answer 8

i got \[\Gamma (4) = 3 \Gamma (3) = 6 \Gamma (2)= 6 \Gamma (1) = 6 . 1! = 6\]

Answer 9

You should get \[\frac{\Gamma(4)}{3^{4}}\]

Answer 10

wait

Answer 11

i'll try again

Answer 12

What integral did you get in terms of u?

Answer 13

First..,
\[\int\limits_{0}^{1} x^{2} \left( \ln \frac{ 1 }{ x } \right)^{2} dx \]
Let x = exp (-u), so dx = = exp (-u) du
Then

\[\int\limits_{0}^{1} (\exp(-u))^{2} \left( \ln \frac{ 1 }{ e^{-u} } \right)^{3} (-e^{-u} du)\]
\[- \int\limits_{0}^{1} e^{-2u} e^{-u} (\ln e^{3})^3 du = - \int\limits_{0}^{1} e^{-3u} (u)^{3} du\]

let y = 3u, so u = y/3, and du = dy/3

Then

\[- \int\limits_{0}^{1} \left( \frac{ y }{ 3 } \right)^{3} e^{-y} \frac{ dy }{ 3 }\]
\[-\frac{ 1 }{ 3 } \int\limits_{0}^{1} \frac{ y^{3} }{ 27 } e^{-y} dy\]
\[-\frac{ 1 }{ 3\times27 } \int\limits_{0}^{1} y^{3} e^{-y} dy\]

Answer 14

so p -1 = 3.., p = 3+1
\[\Gamma (3+1) = 3 \Gamma (3)\]

Answer 15

am I wrong ??

Answer 16

You didn't need the y substitution. You forgot to change your bounds of integration.
From a table I found if a > 0 and n > -1 then we have
\[\int\limits_{0}^{\infty}x^{n}e^{-ax}dx=\frac{\Gamma(n+1)}{a^{n+1}}\]

Answer 17

Oh.., my bad..., thank you for helping me ..,