Mellin Transforms -What could I be doing wrong?

  • Thread starter ~Death~
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In summary, you are using a formula for computing the integral x^(1/4)/(x^2+9)dx from 0 to infinity. You use the convention of polar angles to determine the phase factors of the residues and the two phase factors of the two integrals along the real axis. The sum of the latter two phase factors will be proportional to a sinus term up to some phase factor. If you can precisely write down the residues and the contribution from the two integrals along the real axis, then I can see where you go wrong if you don't get the right answer.
  • #1
~Death~
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I need to compute integral x^(1/4)/(x^2+9)dx from 0 to infinity

so I use a formula

-pie^(-5pi/4i)/sin(5pi4)SUM residues z^(5/4-1)/(z^2+9) so that's sqrt2pie^(-5pi/4i)3^(1/4)/(6i)(e^(pi/8i)-e^(3pi/8i)) (with the 0<arg<2pi branch of sqrt) but e^(-7pi/4)(e^(pi/8i))-e^(3pi/8i)) which is some horrible number...and that's not the answer according to my book

=(
 
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  • #2
Looks like you made a mistake with the phase factors. It is not wise to put in the sin(5 pi/4) factor in there in the way you did.

You choose a convention for the polar angles and that chaoice then determines both the phase factors of the residues and the two phase factors of the two integrals along the real axis. The sum of the latter two phase factors will be proportional to a sinus term up to some phase factor.
 
  • #3
Count Iblis said:
Looks like you made a mistake with the phase factors. It is not wise to put in the sin(5 pi/4) factor in there in the way you did.

You choose a convention for the polar angles and that chaoice then determines both the phase factors of the residues and the two phase factors of the two integrals along the real axis. The sum of the latter two phase factors will be proportional to a sinus term up to some phase factor.

i tried what you said

but i sitill can't seem to find where i went wrong

the formula in the text says -pie^(-pia)/sin(pia) SUM residues of fz^(a-1) excluding 0

= Integral -inf to inf z^(a-1)f(z)
 
  • #4
Just write down the contour, and precisely write down the the residues and the contribution from the two integrals along the real axis, without talking any shortcuts. If you do that, then I can see where you go wrong if you don't get the right answer.
 

What is a Mellin transform?

A Mellin transform is a mathematical operation that converts a function of a single variable into a function of a different variable. It is often used in mathematical analysis and signal processing.

What is the purpose of a Mellin transform?

The main purpose of a Mellin transform is to simplify mathematical operations, particularly in the context of complex functions. It can also help with the analysis of signals and data sets.

How is a Mellin transform different from a Fourier transform?

While both a Mellin transform and a Fourier transform are used to convert functions from one domain to another, they differ in the type of variable they use. A Fourier transform uses complex numbers, while a Mellin transform uses real numbers.

What are some common mistakes made when using Mellin transforms?

One common mistake is not understanding the properties and limitations of the Mellin transform, leading to incorrect applications. Another mistake is not correctly defining the range and domain of the function being transformed.

How can I verify if my Mellin transform is correct?

The best way to verify the accuracy of a Mellin transform is to compare it to known solutions or use numerical methods to evaluate the transformed function. Additionally, checking for the correct properties, such as linearity and shifting, can also help confirm the accuracy of the transform.

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