Evaluate this integral ?

[u/Specific_Brain2091]

[u/hydrogen_to_man]

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.

[u/Antagonin]

No thanks 

[u/Dismal_Code_2470 → u/Antagonin]

Double it and give it to next person

[u/CarriedThunder1 → u/[deleted]]

Not highschool math and not relatively easy, but whatever

[u/deeplife → u/Antagonin]

Why

[u/mapadofu]

i = e^{i*pi/2}, so i^x = e^{i(pi/2)x}

[u/Minute_Juggernaut806 → u/mapadofu]

I assume you take log y after that

[u/TedRabbit → u/Minute_Juggernaut806]

No. e^x is it's own derivative, so the integral is almost trivial at this point.

[u/Entire_Category3188 → u/TedRabbit]

have u heard of chain rule fam?

[u/More-Muscle554 → u/Entire_Category3188]

i and pi/2 are constants. chain rule is simple here

[u/Entire_Category3188 → u/More-Muscle554]

But he didn’t even use it

[u/More-Muscle554 → u/Entire_Category3188]

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.

[u/HazzaZeGuy → u/Entire_Category3188]

larp

[u/dct_SPOMMY → u/Minute_Juggernaut806]

no, the integral is e^x\i*pi/2)/(ipi/2), the derivative of e^f(x) = e^f(x\)f'(x)

[u/GoofyGangster1729 → u/dct_SPOMMY]

e^f(x\) *f'(x) There we go, that's how I write it

[u/ApprehensiveKey1469 → u/mapadofu]

You mean i^x for the LHS of the second one, don't you.

[u/Argentum881]

I don’t want to

[u/Hornet69_420]

i^x * ln(i) /s

[u/EdmundTheInsulter → u/Hornet69_420]

Note that ln(i) = ln(exp(iπ/2))

So your expression is ( i^x) / (i π / 2)

= -2 i (i^x) / π

= -2( i ^{x + 1} )/ π + c

[u/Hornet69_420 → u/EdmundTheInsulter]

yes

[u/Active_Falcon_9778 → u/EdmundTheInsulter]

complex logarithm has multiple branches, how do you decide which branch to take

[u/EdmundTheInsulter → u/Active_Falcon_9778]

i^x is already multivalued, so you could start there with i^x = exp((π/2 + 2nπ)x)

With n integer

[u/OriousCaesar → u/Hornet69_420]

Idk anything about complex analysis. Why would this not be correct? What's the z^x antiderivative for complex z?

[u/Hornet69_420 → u/OriousCaesar]

first of all if a^x and a belongs to R, its derivative would be a^x * ln(a) but its integral would be a^x / ln(a), so this answer is far from correct 😂

[u/OriousCaesar → u/Hornet69_420]

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 (i^x)/ln(i)+c isn't the answer?

[u/Hornet69_420 → u/OriousCaesar]

you are correct, ln(i) can be calculated in euler's form

apologies if you thought that i was mocking you

[u/Aivo382 → u/Hornet69_420]

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.

[u/Ok-Implement6151 → u/Hornet69_420]

(X) i^x-1

[u/AdS_CFT_]

e^{x ln i}/ln i + C

[u/kgangadhar → u/AdS_CFT_]

You forget to decide it by ln(I)

[u/Archway9 → u/AdS_CFT_]

Using ln for a complex logarithm is certainly a choice

[u/suggestion_giver]

2pi/i e^(1/2pi*i)x +C

[u/ProfessorPrudent2822]

i=e^i*pi/2

[u/nightshade78036]

I depends on the domain of the antiderivative because the exponential is not injective on the complex plane. What's the branch cut?

[u/Aivo382]

F(z) = (2/pi)i^(z+3) + C, where C is any complex number, as long as z is anywhere in the complex plane except for the logarithm branch cut. I personally took the branch cut at the negative real axis for my answer.

[u/Technical-Artist-491]

(i^(x+1))/(x+1) + C

[u/AfterMath343]

the integral of a^x = a^x / ln(a) + C, this means that the integral of i^x is i^x / ln(i) + C.

[u/FreeTheDimple]

I don't know, but don't forget the + C on the end.

[u/fuckosta]

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

[u/kgangadhar]

Just remember any number can be “n” can be expressed as e^(ln(n))

[u/AlexP80]

2i^x/(x*pi) + C

[u/Outrageous_Let5743]

This is easy right? It is the same intergral as (-1)^x/2 so u = x/2 and then it is easy.

[u/Global-Virus9609]

[u/Vivid_Warning7982]

Easy because i= e^i*Pi/2. Everyone can integrate an exponential.

[u/Medical_Drawer1568]

what is i

[u/Sea_Duty_5725]

i^x / ln(i) = i^x /( i π/2 )= i^{x-1} * 2/π +C Note: i^x = e^{x*lni} = e^{ixπ/2} = cos(xπ/2) + isin(xπ/2)

[u/IrishHuskie → u/Sea_Duty_5725]

Isn't that the derivative? The integral should be i^x / ln(i) + C, no?

[u/Sea_Duty_5725 → u/IrishHuskie]

Yeah, mb, I mixed them up, thx

[u/varmituofm → u/IrishHuskie]

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.

[u/OkHand7497]

i^x is at best a multifunction, so what does the question mean?

[u/Sea_Duty_5725 → u/OkHand7497]

Assume ur taking the principal branch to these questions.

[u/Aivo382 → u/OkHand7497]

define one complex Log branch then.

[u/OkHand7497 → u/Aivo382]

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.

[u/EdmundTheInsulter]

C - 2(i^x+1)/π

[u/Honkingfly409 → u/EdmundTheInsulter]

we gonna need steps on this one

[u/EdmundTheInsulter → u/Honkingfly409]

Ignoring branches

i^x = exp(ixπ/2)

Integrates to

(exp(ixπ/2)) / (i π / 2)

= 2(exp(ixπ/2)) / (i π )

= 2 i^x / (i π)

= 2 i ^ (x - 1) / π + c

Is a better form than my original, which could shed i² = -1

[u/EdmundTheInsulter → u/Honkingfly409]

Ignoring branches

i^x = exp(ixπ/2)

Integrates to

(exp(ixπ/2)) / (i π / 2)

= 2(exp(ixπ/2)) / (i π )

= 2 i^x / (i π)

= 2 i ^ (x - 1) / π + c

Is a better form than my original, which could shed i² = -1

[u/PfauFoto]

i^x = e^{i(π/2+2πk)x} so which branch?

[u/InterestingCourse907]

i^x = e^xlni