Question

Beta gamma functions.

Answer 1

\[\rm Prove~that~\\ \int\limits_{0}^{\infty}\frac{ \log(i+ax^2) }{x^2 }dx=\pi~\cdot~\sqrt a\]

Answer 2

What's the definition of the beta function that you're used to? There are like 3 ways you can sorta juggle this thing around, and I think usually it comes down to making some substitution to make it in terms of the other.

Answer 3

Here I'll just throw them out here since I could use the practice anyways:

\[\beta(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx \]\[= \int_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}} dy \]\[ =2 \int_0^\frac{\pi}{2} \sin^{2m-1} \theta \cos^{2n-1} \theta d \theta \]\[= \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\]

Since the thing you're wanting to prove has \(\infty\) in the upper bound on the left and \(\pi\) in it on the right I'd probably lean towards trying to move it from that second version to the one with sine or the gamma functions.

Answer 4

well i believe that we can also use DUIS

Answer 5

Awesome I think we should do that first since that's pretty simple to do.

Answer 6

yup!

Answer 7

so dI/da

Answer 8

Yeah that simplifies it a lot. Is i a variable or square root of negative 1?

Answer 9

i believe complex though m not sure hehehe

Answer 10

Haha alright. I've finished DUIS and it looks like it's about a step or two away from being solved.

Answer 11

heheh
but im not able to solve after i get : 1/(i+ax^2)

Answer 12

What I believe is we can make this:
\[\int_0^\infty \frac{dx}{i+ax^2}\]
look like this:
\[\int_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}} dy\]

in order to do that, my guess is we can turn that i in the denominator into a 1 like this:

\[\int_0^\infty \frac{dx}{i+ax^2}=\frac{-i}{-i} \int_0^\infty \frac{dx}{i+ax^2} = -i\int_0^\infty \frac{dx}{1-iax^2}\]

Then try substitution \(-iax^2 = y\) I guess and see what happens

Answer 13

well m not familiar with that second form though
but yea i'll try
please wait

Answer 14

There might be a way around it then, I am just sorta making stuff up as I go here we might not be going towards the answer right now, I'm just tryingt stuff

Answer 15

this is what i got so far :

\[\frac{ -i }{ 2 \sqrt{a} } \int\limits_{0}^{\infty}y^{1-/2}\cdot~(1+y)^{-1}dy\]

Answer 16

@Kainui @ganeshie8

Answer 17

@terenzreignz @texaschic101

Answer 18

let me verify what u got:
\(ax^2 = iy\)
\(2ax dx = i dy\)
\(dx = \dfrac{idy}{2a}\times \dfrac{1}{\sqrt {(iy/a) } } =\dfrac{ \sqrt {(i/a)}}{2\sqrt y}dy\)

I am getting a different constant....

\(\Large =\dfrac{ -i\sqrt i }{ 2 \sqrt{a} } \int\limits_{0}^{\infty}y^{-\frac{1}{2}}\cdot~(1+y)^{-1}dy\)

Anyways, using
\(\Large \int \limits_0^\infty \frac{y^{n-1}}{(1+y)^{m+n}} dy = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\)
n = 1/2 , m = 1/2

\( \int\limits_{0}^{\infty}y^{-\frac{1}{2}}\cdot~(1+y)^{-1}dy =\frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \dfrac{\sqrt \pi \times \sqrt \pi }{1} = \pi \)

Iff we ignore \(-i \sqrt i\) [somewhere there is an error...],
\(f'(a)= \dfrac{\pi }{2\sqrt a} \\ \implies f(a) = \int \dfrac{\pi }{2\sqrt a} da= \pi \sqrt a \)

Answer 19

thanks