Series expansion of integral (ln(x))^2/(1+x^2) dx from 0 to infinity

In summary, the problem asks to evaluate the integral of (ln(x))^2/(1+x^2) from 0 to +infinity using a series expansion of the integrand. By substituting z=e^t and using the even property of the integrand, the integral is simplified to 2 times the integral from 0 to +infinity of (t^2 e^-t)/(1+e^-2t). Expanding this in powers of e^-2t and integrating term by term, we can obtain the desired series representation of the integral.
  • #1
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Hi everyone, once I again I turn to all of your expertise in complex analysis.

Homework Statement


Evaluate

[tex]\int\frac{(ln(x))^{2}}{1+x^{2}}dx[/tex]

from 0 to +infinity by appropriate series expansion of the integrand to obtain

4[tex]\sum(-1)^{n}(2n+1)^{-3}[/tex]

where the sum goes from n=0 to +infinity


Homework Equations



Laurant series, anything in complex analysis

The Attempt at a Solution



At this point, I'm looking for how to even start this. We never covered this in class, and the professor is not really helpful with our (the other students too) questions. I started with the substitution

[tex] z=e^{t} [/tex]

and came up with:

[tex]\frac{1}{2}\int t^{2}sech(t) dt[/tex]

but I think the limits of integration change from -infinity to +infinity with this substitution. either way I am stuck here. I think I have to create a series and integrate term by term. I guess the cleanest way to get to the answer is to leave the series in terms of n and integrate that.

So I guess my question is: should I abandon the substution I have and try to come up with a series of ln about some point other than zero and substitute back to zero at the end? I'm just not sure that's legit.

Either way, just looking for tips. I'm sure it is easy, and it's just a matter of us not learning the material and I can't seem to find this covered in the books I looked at.

Thanks in advance!
 
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  • #2
You're on the right track. Note that the integrand is even, so you can say
$$\int_{-\infty}^\infty \frac{t^2}{e^t+e^{-t}}\,dt = 2\int_0^\infty \frac{t^2}{e^t + e^{-t}}\,dt = 2\int_0^\infty \frac{t^2 e^{-t}}{1 + e^{-2t}}\,dt.$$ Then expand in powers of ##e^{-2t}##. The resulting integrals can be done by parts.
 

What is a series expansion of an integral?

A series expansion of an integral is a method of approximating a definite integral by expressing it as an infinite sum of simpler functions. It is useful for evaluating integrals that cannot be solved through traditional methods.

What is ln(x)?

ln(x) is the natural logarithm function, which is the inverse of the exponential function. It is defined as the power to which the base e (approximately 2.718) must be raised to obtain the number x.

What is the significance of (ln(x))^2/(1+x^2) in this integral?

In this integral, (ln(x))^2/(1+x^2) is the integrand, which is the function being integrated. It is a combination of the natural logarithm function and the reciprocal of a quadratic function.

Why is the integral being evaluated from 0 to infinity?

The integral is evaluated from 0 to infinity because the function (ln(x))^2/(1+x^2) is undefined at x=0 and approaches infinity as x approaches infinity. Therefore, the integral must be evaluated over the entire range to obtain an accurate result.

What are the applications of series expansion of integrals?

The series expansion of integrals has various applications in mathematics, physics, and engineering. It can be used to solve problems involving infinite sums, to approximate functions, and to evaluate integrals that cannot be solved through traditional methods. It is also used in numerical analysis and computer algorithms.

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