#Dimension of Real Improper Integrals(via extension to complex plane)

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novel terrace
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I would like to know if it can be said with certainty and generality that any improper integral
over the real line
of a real function
(said integral converges and exist)
must yield a wholly real value?
It cannot be imaginary nor have imaginary components.

Given an improper integral from 0 to infinity, i suppose we can't use jordan's lemma(I am still working on learning the proof) directly.

So my thought then is to use a quartercircle contour.

But 1st, let me just change it complex plane. change all x to z.

I get improper integral[0, inf] ((2z^2)/(z^4 + 1))

Working to get the poles:
(z^4 +1) = (z^2 + i)(z^2 -i) = (z + sqrt(i))(z - sqrt(i))(z + i sqrt(i))(z - i sqrt(i))

These are simple poles of order 1.

From JL, i can work out that the quarter circle curve approaches 0, as R, from z = Re^itheta, approaches infinity.

I can do residue theorem. The quarter circle only encloses z = sqrt(i)

Res(f, sqrt(i) = (2(i))/((2 sqrt(i))(i+ i) = 1/2(sqrt(i)
multiply that by 2pii, we get pi* (i/sqrt(i))

So, after alot of days(hence my id), i thought about changing it to theta form and then expanding it in polar? form, and i get

pi(cos(pi/4) + i sin(pi/4)) = Improper integral over the positive real axis + improper integral over the positive part of imaginary axis

The issue is the contribution along the imaginary axis might not be trivial.

(Funnily enough, i try to parameterise the imaginary component, and i end up something like the real improper integral, except with i with parameterising variable t and its "end points" being completely real valued, but i will not assume)

Which leads to my question: Are improper integrals "limited" by the dimension(? for my lack of better vocabulary)?
like, an improper integral over the real line for a real function.
Even if we solve it via extending to complex plane, the answer MUST be real, not imaginary, not complex.

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radiant flint
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if you do contour integral over C you still take the real part at the end

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so the result is a real number if the improper integral converges

novel terrace
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i did a contour integral, let C be the contour

  • the positive real line from 0 to +inf
  • the arc/etc over the complex part(that can be shown to vanish in JL)
  • the positive part of the imaginary axis from +i inf to 0

my result is a pi(cos(pi/4) + i sin(pi/4))
we know pi cos(pi/4) is real
and pi i sin(pi/4)) is imaginary.

novel terrace
radiant flint
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yes of course

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the same intuition still applies: the integrand is a real-valued function and we ask whether the "area" under the curve is finite, so it goes without saying that the result is a real number

novel terrace
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wait isn't "finite" a question of big or small(to the point of converges or not) while real/imaginary is which plane it lies on.

I mean, i could get the "simple intuition" like what borns from real should, gets thrown with real operations, has to end up real

novel terrace
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@radiant flint well. thank you! imma close this~

gleaming copperBOT
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@novel terrace has given 1 rep to @radiant flint

novel terrace
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+close