Feynman's integration trick | Mathematics | Page 1

split aspen Apr 10, 2023, 1:21 AM

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Where did I go wrong?

stuck ravenBOT Apr 10, 2023, 1:21 AM

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split aspen Where did I go wrong?

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Wait patiently for a helper to come along.
Once someone helps you, say thank you and close the thread with: ```diff
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split aspen Apr 10, 2023, 1:23 AM

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This integral is equal to π/4 but I'm getting -3π/4

fading saddle Apr 10, 2023, 3:04 AM

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Limits wrong by the end, integral you want is $\int_0^\infty \frac{1}{(t+1)^2+1} , dt$

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that's all I'm giving you because your pencil's tone hurts my eyes

mellow pierBOT Apr 10, 2023, 3:09 AM

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agex

split aspen Apr 10, 2023, 3:22 AM

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fading saddle Limits wrong by the end, integral you want is $\int_0^\infty \frac{1}{(t+1)^2+1...

Then on LHS, I(t) as t goes to infinity diverges to infinity due to $x^t$ term

mellow pierBOT Apr 10, 2023, 3:22 AM

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∫ i dx

split aspen Apr 10, 2023, 3:22 AM

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fading saddle that's all I'm giving you because your pencil's tone hurts my eyes

Sorry about that

fading saddle Apr 10, 2023, 3:23 AM

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what are you talking about, that integral is π/4

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you want that integral by the end

split aspen Apr 10, 2023, 3:24 AM

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fading saddle what are you talking about, that integral is π/4

But then how do you get the original integral I?

fading saddle Apr 10, 2023, 3:25 AM

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your I'(t) by the 5th till last line is wrong

split aspen Apr 10, 2023, 3:30 AM

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Can you please tell exactly what went wrong in it cause I am not getting?

fading saddle Apr 10, 2023, 3:34 AM

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try again with this:
$I(t)=\int_0^1 \frac{x^{it}-x^{-i}}{2i\ln{x}} , dx$

mellow pierBOT Apr 10, 2023, 3:34 AM

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agex

fading saddle Apr 10, 2023, 3:34 AM

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using the complex sine definition earlier on than you did

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almost the same overall method

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I(1)=I

split aspen Apr 10, 2023, 3:38 AM

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I got the answer using $I(t) = \int_0^1 \frac{sin(tlnx)}{lnx}, dx$

mellow pierBOT Apr 10, 2023, 3:38 AM

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∫ i dx

split aspen Apr 10, 2023, 3:39 AM

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But wanted to why was this method producing wrong answer

fading saddle Apr 10, 2023, 3:44 AM

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fading saddle your I'(t) by the 5th till last line is wrong

something happened just before here, but it's almost 5am so icba to actually pinpoint it

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$I'(t)=\int_0^1 \sin(\ln x)x^t, dx$ then if you just let everything be and remember to resubstitute later:
$\frac{x^{t+1}((t+1)\sin(\ln x)-\cos(\ln x))}{t^2+2t+2}$

mellow pierBOT Apr 10, 2023, 3:53 AM

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agex

fading saddle Apr 10, 2023, 3:53 AM

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bounds still 0→1

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and that gets $-\frac{1}{t^2+2t+2}$

mellow pierBOT Apr 10, 2023, 3:54 AM

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agex

fading saddle Apr 10, 2023, 3:55 AM

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as our I'(t)

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then we integrate relative to t -arctan(t+1)

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and resubstitute our 0

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where'd that pesky negative come from

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hahayes

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don't care, area is positive

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π/4

split aspen Apr 10, 2023, 4:10 AM

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How can you do that? despair

fading saddle Apr 10, 2023, 4:12 AM

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do what

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@split aspen

split aspen Apr 10, 2023, 5:07 AM

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fading saddle do what

Ignore the - and forget the constant

fading saddle Apr 10, 2023, 5:07 AM

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split aspen Ignore the - and forget the constant

didn't forget the constant, also you can make it positive if you plot a couple points

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the graph is almost strictly positive on 0,1

fading saddle Apr 10, 2023, 5:08 AM

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fading saddle didn't forget the constant, also you can make it positive if you plot a couple p...

I made the constant 0 because that's what it started as

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before I took the derivative

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agex's integration trick:

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measure theory

split aspen Apr 10, 2023, 5:11 AM

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fading saddle I made the constant 0 because that's what it started as

But then take lim t-> -infinity I(t) with original definition is 0, but this new definition with arctan gives π/2

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So constant must be -π/2 how 0?

fading saddle Apr 10, 2023, 5:11 AM

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split aspen But then take lim t-> -infinity I(t) with original definition is 0, but this new...

that's not what the constant was before you took the derivative

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there was no constant before you took the derivative

split aspen Apr 10, 2023, 5:12 AM

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Does that matter?

fading saddle Apr 10, 2023, 5:12 AM

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yes

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because that's what the constant is

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you get the constant from indefinite integration because it's reversing differentiation

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which removes constants

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if you didn't have any constants in the first place and you know that, then you have a condition

split aspen Apr 10, 2023, 5:13 AM

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But then what about two different values of I(t) at -infinity?

fading saddle Apr 10, 2023, 5:13 AM

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so you can make C=0

fading saddle Apr 10, 2023, 5:13 AM

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split aspen But then what about two different values of I(t) at -infinity?

my version doesn't need limits 😏

split aspen Apr 10, 2023, 5:14 AM

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I am not getting this 😭

fading saddle Apr 10, 2023, 5:14 AM

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I didn't change the bounds because I resubbed everything back in

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simplifies with redundancy

split aspen Apr 10, 2023, 5:15 AM

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Sorry but I still don't get it

fading saddle Apr 10, 2023, 5:16 AM

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ok

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have you done inverse trig substitution and hyperbolic trig substitution yet?

split aspen Apr 10, 2023, 5:16 AM

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Not hyperbolic sub

fading saddle Apr 10, 2023, 5:17 AM

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why are you doing Feynman's method and complex substitution then weep

split aspen Apr 10, 2023, 5:18 AM

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I liked Feynman's method and was able to evaluate many cool integrals with it until I happened to step on this integral

fading saddle Apr 10, 2023, 5:18 AM

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lmao

split aspen Apr 10, 2023, 5:18 AM

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I haven't done any official course in calculus I just learnt from internet

fading saddle Apr 10, 2023, 5:18 AM

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if you really want to redo your method and check each step then check everything up until the 5-6th till last line

split aspen Apr 10, 2023, 5:19 AM

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I did check many times but don't know what went wrong

fading saddle Apr 10, 2023, 5:19 AM

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ok

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my guess is it's something to do with your complex substitution

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but your page and lead blinds me

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so I can't say for sure

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don't even need complex substitution to evaluate that integral

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just your lnx sub and then it's fairly simple integration by parts twice

split aspen Apr 10, 2023, 5:22 AM

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Okay thanks I'll try that

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+close

#Feynman's integration trick