Adding restrictive conditions to a limit calculation

[MathematicalPhysicist]

I have the following limit to calculate under the assumption that $\Re(x+y)>1$:

Limit[Integrate[
  1/((1 + t^2)^n*(1 + I*t)^x (1 - I*t)^y), {t, -Infinity, Infinity}],
n -> Infinity]

I want to add the above assumption for integral, how to do it? does it even converge?

Thanks.

 

[mmwave]

I can provide some guidance on how to approach this limit under the given assumption. First, we can rewrite the integral as follows:

Limit[Integrate[1/((1 + t^2)^n*(1 + I*t)^x*(1 - I*t)^y), {t, -Infinity, Infinity}], n -> Infinity] = Limit[Integrate[1/((1 + t^2)^n*(1 + I*t)^x*(1 - I*t)^y), {t, -Infinity, Infinity}, Assumptions -> Re[x + y] > 1], n -> Infinity]

This will add the assumption that $\Re(x+y)>1$ to the integral. Now, to determine if the integral converges, we can use the Cauchy's convergence test. This test states that if the absolute value of the integrand decreases as t increases, the integral will converge. In this case, the integrand is 1/((1 + t^2)^n*(1 + I*t)^x*(1 - I*t)^y), and we can see that the denominator will increase as t increases, while the numerator is a constant. Therefore, the integral will converge if $\Re(x+y)>1$.

To determine the exact value of the limit, we can use the Method of Residues. This method involves finding the residues of the integrand at its singularities (in this case, at t = ±i). The residues can then be used to evaluate the integral. However, this method can be quite complex and may require advanced mathematical techniques.

In conclusion, the given limit will converge if $\Re(x+y)>1$, and the exact value can be determined using the Method of Residues. I hope this helps!