Question

beta gamma functions :

Answer 1

\[\int\limits_{0}^{\infty}(\sqrt{x}~\cdot~e^{-x^2}dx)\int\limits_{0}^{\infty}\frac{ e^{-x^2} }{\sqrt{x}}dx\]

Answer 2

multiplication of two integrals so how do i go about?

Answer 3

since i know the substitution

Answer 4

so should i solve the one of the integral first n then multiply ?

Answer 5

let x^2=t

Answer 6

\[\int\limits_{0}^{\infty}(\sqrt{x}~\cdot~e^{-x^2}dx)\int\limits_{0}^{\infty}\frac{ e^{-x^2} }{\sqrt{x}}dx = \int\limits_{0}^{\infty}\int\limits_{0}^{\infty}\sqrt{\frac{x}{y}}~\cdot~e^{-(x^2+y^2)}dx\, dy\]

Answer 7

'.'
how did u do?
u took x=y?

Answer 8

please elaborate Sir

Answer 9

scratch that, you want to express it in terms of gamma function right ?
then x^2 = t looks good to me!

Answer 10

okay
after that i substituted it into both integrals
so now how do i solve further?
as separate integrals or..?

Answer 11

please answer :3

Answer 12

\(x^2 = t \\ \implies 2xdx = dt \\\implies dx = \dfrac{1}{2}t^{-1/2}dt\)

\[\int\limits_{0}^{\infty}\frac{ e^{-x^2} }{\sqrt{x}}dx \\~\\=\int\limits_{0}^{\infty} \dfrac{e^{-t}}{t^{1/4}}\, \dfrac{1}{2}t^{-1/2}dt \\~\\= \dfrac{1}{2} \int\limits_0^{\infty}t^{-3/4}e^{-t}\,dt \\~\\ = \dfrac{1}{2} \int\limits_0^{\infty}t^{-1/4-1}e^{-t}\,dt \\~\\=\dfrac{1}{2}\Gamma(1/4) \]

Answer 13

As you can see, its just algebraic manipulation. You just need to try and put the given integral in the gamma integral form :

https://upload.wikimedia.org/math/6/0/3/603515ff4df2cb431387e92bb6419c66.png

Answer 14

And yes, doing them separately seems to work..

Answer 15

i got the same as you got

Answer 16

wait i got t^{3/4}

Answer 17

looks i have got t^{-3/4} ?

Answer 18

oops sorry
yea its 1/4

Answer 19

How about the other integral ?

Answer 20

please wait

Answer 21

same t^{1/4}

Answer 22

I'm not following you, if psble could you take a screenshot of your work and attach ?

Answer 23

wait
i got the same as the above work done by u

Answer 24

Okay good, what about the first integral ?

Answer 25

so its getting squared

Answer 26

\[\int\limits_{0}^{\infty}(\sqrt{x}~\cdot~e^{-x^2}dx) = ?\]

Answer 27

both turn out to be the same

Answer 28

wait
ill write it down

Answer 29

I'm getting \[\int\limits_{0}^{\infty}(\sqrt{x}~\cdot~e^{-x^2}dx) = \dfrac{1}{2}\Gamma(3/4)\]

Answer 30

Just double check your work..

Answer 31

oops lol
ROFL i was looking at the same integral!!!

Answer 32

yup yup it is 3/4
so shall i leave it like that?

Answer 33

@ganeshie8
ans=\(\huge\frac{\sqrt2\cdot \pi}{4}\)

Answer 34

Oh that looks nice, lets try and simplify maybe

Answer 35

i believe this will also work
usually we keep it till this ans

Answer 36

\[\int\limits_{0}^{\infty}(\sqrt{x}~\cdot~e^{-x^2}dx)\int\limits_{0}^{\infty}\frac{ e^{-x^2} }{\sqrt{x}}dx = \dfrac{1}{2}\Gamma(3/4)*\dfrac{1}{2}\Gamma(1/4)\]

Answer 37

thats the final ans i got

Answer 38

Familiar with below relation between beta and gamma functions ?
\[\beta(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\]

Answer 39

Rearranging above identity a bit we get
\[\Gamma(a)\Gamma(b) = \Gamma(a+b)*\beta(a,b)\]

Answer 40

try using that and see if it simplifies any further..

Answer 41

i used another property :)

Answer 42

Yeah I have the feeling that there must be some smart way to work this... What property have you used ?

Answer 43

\(\Gamma(n) \Gamma(1-n)=\Large \frac{\pi}{sin(n\pi)} \\ where 0<n<1\)

oops the ans it should be sin(3\(\pi\)/4)