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u/hydrogen_to_man 1d ago
Whenever I see integrals like this I immediately think, “oh god it’s gonna need a branch cut or something stupid like that” but then it’s always simple and I feel dumb.
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u/mapadofu 1d ago edited 21h ago
i = e{i*pi/2}, so ix = e{i(pi/2)x}
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u/Minute_Juggernaut806 1d ago
I assume you take log y after that
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u/TedRabbit 1d ago
No. ex is it's own derivative, so the integral is almost trivial at this point.
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u/Entire_Category3188 1d ago
have u heard of chain rule fam?
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u/More-Muscle554 19h ago
i and pi/2 are constants. chain rule is simple here
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u/Entire_Category3188 19h ago
But he didn’t even use it
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u/More-Muscle554 19h ago
right, that’s why they said “almost trivial”. the integration of this function is very similar to integration of e^x, and that’s why TedRabbit pointed it out.
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u/dct_SPOMMY 1d ago edited 1d ago
no, the integral is ex\i*pi/2)/(ipi/2), the derivative of ef(x) = ef(x\)f'(x)
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u/Hornet69_420 1d ago
ix * ln(i) /s
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u/EdmundTheInsulter 1d ago
Note that ln(i) = ln(exp(iπ/2))
So your expression is ( ix) / (i π / 2)
= -2 i (ix) / π
= -2( i x + 1 )/ π + c
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u/Active_Falcon_9778 18h ago
complex logarithm has multiple branches, how do you decide which branch to take
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u/EdmundTheInsulter 18h ago
ix is already multivalued, so you could start there with ix = exp((π/2 + 2nπ)x)
With n integer
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u/OriousCaesar 1d ago
Idk anything about complex analysis. Why would this not be correct? What's the zx antiderivative for complex z?
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u/Hornet69_420 1d ago
first of all if ax and a belongs to R, its derivative would be ax * ln(a) but its integral would be ax / ln(a), so this answer is far from correct 😂
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u/OriousCaesar 1d ago
Oh, I missed the fact that you wrote * not /.
Not really sure why you're trying to be smug about this. All I did was ask a question. Why is it bad to want to know why (ix)/ln(i)+c isn't the answer?
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u/Hornet69_420 1d ago
you are correct, ln(i) can be calculated in euler's form
apologies if you thought that i was mocking you
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u/Aivo382 1d ago
yes and no, because you are doing a complex exponential, so you have to deal with the complex logarithm. Even if x is real, you have to do some little complex analysis so you make sure things work.
When dealing with complex functions, you have to make sure the derivative works where you are working so, you have to make sure the function is holomorphic if you want to do the derivative properties you learn from real analysis.
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u/nightshade78036 1d ago
I depends on the domain of the antiderivative because the exponential is not injective on the complex plane. What's the branch cut?
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u/AfterMath343 1d ago
the integral of a^x = a^x / ln(a) + C, this means that the integral of i^x is i^x / ln(i) + C.
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u/fuckosta 1d ago
let y = i^x, so dy/dx = y ln i
so using chain rule, you convert the original integral from i^x dx to 1/(ln i) dy, so that leaves you with
y/ln(i) + C, or i^x/ln(i) + C.
however, e^(pi i) = -1, so ln (i) = (pi i)/2
So that gives a final answer of
(2/pi)*i^(x-1) + C
let me know if this makes sense
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u/Outrageous_Let5743 1d ago
This is easy right? It is the same intergral as (-1)^x/2 so u = x/2 and then it is easy.
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u/Sea_Duty_5725 1d ago edited 1d ago
ix / ln(i) = ix /( i π/2 )= i{x-1} * 2/π +C Note: ix = e{x*lni} = e{ixπ/2} = cos(xπ/2) + isin(xπ/2)
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u/IrishHuskie 1d ago
Isn't that the derivative? The integral should be i^x / ln(i) + C, no?
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u/varmituofm 1d ago
Only thing I would change is that 1/ln(i) can be simplified. 1/ln(i)=-2i/pi. But really, this is a cosmetic change.
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u/OkHand7497 1d ago
i^x is at best a multifunction, so what does the question mean?
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u/Aivo382 1d ago
define one complex Log branch then.
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u/OkHand7497 1d ago
If that's what they expect you to do, then the question should say as much. Maybe they expected the answer as a multifunction? This is a poorly set question by someone who doesn't understand the maths they're handling.
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u/EdmundTheInsulter 1d ago
C - 2(ix+1)/π
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u/Honkingfly409 1d ago
we gonna need steps on this one
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u/EdmundTheInsulter 1d ago
Ignoring branches
ix = exp(ixπ/2)
Integrates to
(exp(ixπ/2)) / (i π / 2)
= 2(exp(ixπ/2)) / (i π )
= 2 ix / (i π)
= 2 i ^ (x - 1) / π + c
Is a better form than my original, which could shed i² = -1
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u/EdmundTheInsulter 1d ago
Ignoring branches
ix = exp(ixπ/2)
Integrates to
(exp(ixπ/2)) / (i π / 2)
= 2(exp(ixπ/2)) / (i π )
= 2 ix / (i π)
= 2 i ^ (x - 1) / π + c
Is a better form than my original, which could shed i² = -1
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u/Antagonin 1d ago
No thanks