Integral Question

[obtuse zephyr]

How do I find I(1) and I(3/2) with the given data.
Don’t know what to do after that.
****(Int is from 0 to infinity)

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[obtuse zephyr]

How does the given data help in solving it?

Don’t we have to evaluate one of the integral for solving it?

[placid veldt]

What do you need to find I’m confused

Just know that if G is the gamma function then G(x) * G(1-x)= pi/sin(pi*x)

Oh is I the gamma function

?

No never mind it’s from -infinity to infinity

[obtuse zephyr]

I don’t know that much about gamma function

This question came in my comprehension with the given data and data is said to be sufficient to solve the given question but I still can’t do it.

[obtuse zephyr]

I don’t think you have to know a whole lot about gamma function to solve it as we haven’t learned about it till now.

[placid veldt]

Okay so do you have trouble finding A . B?

[obtuse zephyr]

Yes

As A is I(1) and B is I(3/2) for those I need I(0) and I(-1/2)

[placid veldt]

Oh okay I see

It’s a bit odd though because the integral itself doesn’t necessarily converge if it was from 0 to infinity there wouldn’t be a problem for p>0 but when t is negative -t is positive

Are you sure it isn’t from 0 to infinity the bounds ?

And that’s how the gamma function is initially defined so seems odd that the bounds are -infinity infinity instead of 0 infinity

Plus you are integrating the power of a negative like if t<0 and s=1/2 there is a problem

[obtuse zephyr]

I don’t remember correctly

Try doing it from 0 to infinity

But if I take from 0 to infinity then will you find I(0) ?

Or are they already known?

[placid veldt]

If it’s indeed from 0 to infinity then I(0) doesn’t exist because the integral diverges

And if it’s indeed from 0 to infinity you can find A and B using a change of variables

Setting x^4=u you will have dx=1/4 * u^(1/4-1) du

so A becomes the integral of 1/4 * u^(1/4-1) * e^-u

Which is 1/4 *I(1/4)

[obtuse zephyr]

Then the other would become I(5/8)?

*1/4 as well

[placid veldt]

Yes the other one is 1/4 * I(3/4)

And 3/4=1-1/4

So using the fact that I(s) * I(1-s)=pi/sin(spi) you’ll get the answer

[obtuse zephyr]

It isn’t I(3/4)

Wait let me recheck

[placid veldt]

Well using the same change of variables

x^2 becomes u^1/2

Multiplying by 1/4 * u^(1/4-1) gives u^-1/4

[obtuse zephyr]

So it becomes u^1/2/u^3/4

Which is u^-1/2

Which is I(3/2)

[placid veldt]

well it’s u^1/2 * u*(1/4-1) which is u^-1/4 which is u^(3/4-1)

[obtuse zephyr]

Oh sorry

[placid veldt]

So B= 1/4 * I(3/4)

[obtuse zephyr]

Yeah thanks mate

[placid veldt]

Np

[obtuse zephyr]

I didn’t change the denomination while doing it

I have some other questions can you help?

[placid veldt]

Im a bit occupied at the moment but you can still post the questions here if you want

[obtuse zephyr]

Alright

Can we find a general answer to alphaAn+2=(beta)*An+1+ (gamma)*An?

[placid veldt]

If (An) is any sequence then yes you can actually

Given alpha is nonzero

[obtuse zephyr]

An is coefficient of x^n in the sequence

[placid veldt]

Okay I see

[obtuse zephyr]

How to approach these kind of questions?

A headway would be great

[placid veldt]

Well maybe you can try identifying a11 using the relation between the roots and the coefficients here the roots are j and j^2 and both have multiplicity 20 so you can possibly find all the coefficients ( here j is the complex number e^2pi *i/3 ) another possibility is using Taylor’s formula to find ak for any k ( basically ak=P^(k) (0)/k! With « P » denoting the polynomial here, so you can calculate the k th derivative of P at 0 using Leibniz’s formula P=(X-j)^20 * (X-j^2)^20 which is equivalent to the root formula I think

Hmmm maybe it’s a bit too much

Here you don’t need the coefficients necessarily but you need to find (alpha-beta)/gamma

Is that right ?

[obtuse zephyr]

Haven’t studied Leibniz formula

[obtuse zephyr]

Yes

[placid veldt]

Okay never mind then

You can try binomial expansion but there will be a double sum

Im sorry I’m quite tired so I said some quite useless stuff

[obtuse zephyr]

Binomial expansion isn’t the way as it’s too length and calculative

I tried doing that In the paper

[placid veldt]

Yes that’s what I thought there is a double sum so it’s quite annoying

[obtuse zephyr]

No need to help me rn you can come whenever you are free

Appreciate the help bro thanks