Question

Evaluate
\[\int\limits_{0}^{1}[{{1-x^2 \over 1+x^2}]^{1/2}}dx\]
In terms of the Gamma function

Answer 1

hello myininaya!

Answer 2

hope you had a good holiday.

Answer 3

hmm

Answer 4

does it involve trig substitution

Answer 5

could you please tell me in which chapter this in calc

Answer 6

when i tried trig substitution
1-x^2 = cos^2(x)
1+x^2 = sec^2(x)

i get
\[\int\limits_{0}^{1}\cos^2(x) dx\]

Answer 7

use the half angle formula

Answer 8

subject; differential equations
In my book the chapter is; Special Functions;
the topic is; beta and gamma functions

Answer 9

i can not see where to apply the half angle formula salem123

Answer 10

cos^2x=1/2(1+cos2x)

Answer 11

do u think this works?

Answer 12

i dont know

Answer 13

but how did you end up only with cos(x)^2

Answer 14

how did you get rid off sec(x)^2

Answer 15

well √cos^2(x)/sec^2(x) = √ cos^4(x) = cos^2(x)

Answer 16

right
?

Answer 17

okay i see

Answer 18

right

Answer 19

if this is right then the half angle formula should be applicable then

Answer 20

but the i am looking how to express it in term of gamma

Answer 21

by the way i have the answer

\[={√π \over 4} \times ({\Gamma(1/4) \over \Gamma(3/4)}-4{\Gamma(3/4) \over \Gamma(1/4)})\]

Answer 22

you want to know how to get there

Answer 23

yeapp

Answer 24

no idea to be honest i can't help

Answer 25

i am taking differential equation but we haven't taken this stuff
yet

Answer 26

thanks any way

Answer 27

This term \( \sqrt{\pi} /4 \) is a multiple of \( \Gamma(1/2) \).

This is an interesting question; I'll think about it.

Answer 28

...tomorrow though. Time for bed to digest all the pie and turkey!

Answer 29

i gotta go to , yeah i noticed the Gamma of (1/2) but couldn't reverse engineer the problem correctly

Answer 30

working backwards from \[=√π/4×({Γ(1/4) \over Γ(3/4)}−4{Γ(3/4) \ \over Γ(1/4)})\]
\[=1/4[{ Γ(1/2)(Γ(1/4) \over Γ(3/4)}−4{Γ(1/2)Γ(3/4) \over Γ(1/4)}]\]
\[= {B_{(1/2,1/4)}\over2} + B_{(-1/2, 3/4)} \]

Answer 31

Never mind. I see now, \( -2\Gamma(1/2) = \Gamma(-1/2) \)

Answer 32

\[Γ(z) = Γ(z+1)/z\]
\[Γ(1/2) =-1/2 Γ(-1/2)\]

Answer 33

yeah

Answer 34

Ok. So this is now doable: show the integral is equal to the sum of those two beta functions.

Answer 35

btw, isn't \( B(x,y) = \frac{ \Gamma(x) \Gamma(y)}{\Gamma(x+y)} \), so have you got the constants quite right?

Answer 36

i think i have it right
(i havent included all the steps, but wolfram can confirmed equivalence)

Answer 37

screw Wolfram; it's not infallible. You need to know for yourself that this is right.

\[ \Gamma(1/2) = \sqrt{\pi} \] so I think the constants aren't quite right.

Answer 38

oh wait
i mean
\[=1/4 \times B(1/2,1/4)+1/2 \times B(−1/2,3/4)\]

Answer 39

(off by a factor of 2)

Answer 40

Right.

Answer 41

so
\[\int\limits_{0}^{1}\sqrt {(1-x^2) \over (1+x^2)})dx={1 \over 2}({B_{(1/2,1/4)} \over 2} + B_{(-1/2,3/4)})\]

Answer 42

\[=\int\limits_{0}^{1}(1-x^2)^{1/2}(1+x^2)^{-1/2}dx\]