Integrating with indented contour

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SUMMARY

The integral \(\int_{-\infty}^\infty \frac{dx}{x(x^2+x+1)}\) does not converge in the Riemann sense due to a pole on the path of integration. However, it can converge in the Cauchy Principal Value sense, as indicated by the discussion. The use of the definite real integral widget on Wolfram confirmed the divergence in the Riemann sense. To evaluate this integral correctly, one should apply the Principal Value method using the command: Integrate[1/(x*(x^2 + x + 1)), {x, -Infinity, Infinity}, PrincipalValue -> True.

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Homework Statement


Evaluate the following integral by integrating the corresponding complex function.

\int_{-\infty}^\infty \frac{dx}{x(x^2+x+1)}

Homework Equations



Cauchy's Residue Theorem for simple pole at a:Res(f;a)=\displaystyle\lim_{z\rightarrow a} (z-a)f(z)

The Attempt at a Solution


I have used the definite real integral widget on wolfram which states that the integral does not converge. Will I be able to show this is the case by integrating around the semi circular contour indented at 0?
 
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jimmycricket said:

Homework Statement


Evaluate the following integral by integrating the corresponding complex function.

\int_{-\infty}^\infty \frac{dx}{x(x^2+x+1)}

Homework Equations



Cauchy's Residue Theorem for simple pole at a:Res(f;a)=\displaystyle\lim_{z\rightarrow a} (z-a)f(z)

The Attempt at a Solution


I have used the definite real integral widget on wolfram which states that the integral does not converge. Will I be able to show this is the case by integrating around the semi circular contour indented at 0?

Did you look at it first? I mean plot it say from -10 to 10? Looks to me it has the right shape to converge in the Cauchy Principal Value sense. That is the only way it can converge since it has a pole on the path of integration. Alpha is telling you it diverges in the Riemann sense. Did you try:

Code:
Integrate[1/(x*(x^2 + x + 1)), 
  {x, -Infinity, Infinity}, PrincipalValue -> True]

However, if you're not familiar with Principal-valued integrals, you may want to look that up.

Edit: made a mistake with the function. It's +1 and I corrected it above but still everything I said in regards to the function with -1 applies to this function as well.
 
Last edited:

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