Solve Challenging Integral with Proven Techniques | x>1 Integer Solution

[pkmpad]

Hello.

I am having a lot of trouble trying to solve/analyse this integral:

$$\displaystyle \int_2^\infty \frac{x+y}{(y)(y^2-1)(\ln(x+y))} dy$$

I have tried everything with no result; it seems impossible for me to work with that natural logarithn.

I have also tried to compute it, as it converges for positive values of x, but that does not help neither.

It is given that that x will be an integer x>1.

Is there any way to leave the integral in terms of x? And to get its asymptotic behaviour?

Thank you very much.

 

[Svein]

Try splitting the integrand into three or four parts: $\frac{x+y}{y(y^{2}-1)\ln(x+y)}=\frac{A}{y}+\frac{B}{y-1}+\frac{C}{y+1}+\frac{D}{\ln(x+y)}$.

 

[pkmpad]

Try splitting the integrand into three or four parts: $\frac{x+y}{y(y^{2}-1)\ln(x+y)}=\frac{A}{y}+\frac{B}{y-1}+\frac{C}{y+1}+\frac{D}{\ln(x+y)}$.

Thank you for your interest.

If I tried that, wouldn't I have a sum of divergent integrals?

 

[Svein]

Thank you for your interest.

If I tried that, wouldn't I have a sum of divergent integrals?

I do not know, but since the lower integral limit is 2 and x is greater than 1 you should be safe.

 

[pkmpad]

I do not know, but since the lower integral limit is 2 and x is greater than 1 you should be safe.

Breaking it in 4 fractions would definitively lead to a sum of divergent integrals, but dividing it into 2 integrals:
$$x \displaystyle \int_2^\infty \frac{1}{y(y^2-1)\log(x+y)} + \int_2^\infty \frac{1}{(y^2-1)\log(x+y)}$$

Since the value of the second one is too small for large x, any idea about what to do with the first one?

 

[Svein]

I am speculating about the complex domain and the residue theorem, but as of now I have not figured out exactly how to apply them.

 

[Svein]

I can give some sort of an answer, but not very useful.

1. Substitute p for x and z for y (just in order not to get confused).
2. Assume p ≥ 3
3. The integrand now reads $\frac{z+p}{z(z-1)(z+1)log(z+p)}$
4. Calculate the residues at z = -1, 0, 1, 1-p:

$Res_{z=-1}:\frac{p-1}{(-1)(-2)\log(p-1)}=\frac{p-1}{2+\log(p-1)}$
$Res_{z=0}:\frac{p}{1(-1)\log(p)}=\frac{-p}{\log(p)}$
$Res_{z=1}:\frac{p+1}{1\cdot 2\log(p+1)}=\frac{p+1}{2 \log(p+1)}$
$Res_{z=1-p}:\frac{1}{(1-p)(2-p)(-p)}=\frac{-1}{p(1-p)(2-p)}$

Given that, we can find the primary Cauchy value of $\int_{-\infty}^{\infty}\frac{z+p}{z(z-1)(z+1)log(z+p)}dz$ from the sum of the residues times πi:
$$\pi i (\frac{-p}{\log(p)}+\frac{p-1}{2\log(p-1)}+\frac{p+1}{2\log(p+1)}-\frac{1}{p(1-p)(2-p)})$$
or, rewritten with x instead of p:
$$\pi i (\frac{-x}{\log(x)}+\frac{x-1}{2\log(x-1)}+\frac{x+1}{2\log(x+1)}-\frac{1}{x(1-x)(2-x)})$$
Given that x should be an integer greater than 1, we have excluded the possible value of x = 2, where we get a (sort of) double pole at -1.

 

[pkmpad]

Thank you! That helps a lot in my problem in spite of being focused on the complex domain of the function