Supreme Integral with Feynman's Trick

We will do the integral of sin(ln(x))/ln(x) from 0 to 1 by using Feynman's Trick (aka differentiation under the integral sign). This is also closely related to the Dirichlet Integral, which is the integral of sin(x)/x from 0 to inf.
For the complex exponential definition of sine 👉 • Complex definitions of...
For another integral by Feynman's Technique: 👉 • Feynman technique: int...
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[@JoJoJet100]

I REALLY like it when you do improper integrals. It's so much more satisfying to get an answer that is an actual number instead of a bunch of math functions added together.

[@blackpenredpen]

: ) I actually like indefinite integral more tho

[@ripjawsquad]

@@blackpenredpen fr

[@atheybengala5720]

@@blackpenredpenfr

[@yoylecake313]

@@blackpenredpen fr

[@ali-zl9ls]

@@blackpenredpenfr

[@jlxip]

17:20 "So, this right here is pretty much the answer but what the heck in the world is this?" I'm crying 😂

[@SteamPunkLV]

from now on, I'll write that instead of pi/4 😂

[@dijkstra4678]

math video: so basically that's the answer. me: ok but what the heck in the world is this?

[@mortadhaalaa5907]

I did it quite peacefully using feynman's trick with the parametrization: I(t) = sin(t lnx) / lnx I guess I've been watching too much flammable maths vids 😂 awesome video nonetheless 🍫

[@hetsmiecht1029]

I think your solution is more elegant, as it doesn't require complex numbers inside natural logs (which can have infinitely many values).

[@euva209]

Nice! It leads to dI/dt = (1/t) ∫e^(u/t)cos(u)du from -∞ to 0 = 1/(t²+1); After integrating you get I = arctan(t) + C; I(0) =0 = C ; I(1) = π/4

[@GauravG91]

Same bro.. And it doesn't involve complex numbers in any way so simple, but a little longer..

[@KewlWIS]

idk how it works, you would have to evaluate it from 0 to 1, you cant have 0 inside the ln tho even if u say t = 0 the integral has a 1/t after solving so can you explain please?

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

@@KewlWIS I = {0,1}∫ sin(ln x)/ ln x dx F(t) = {0,1}∫ sin(t*ln x) / ln x dx => F'(t) = {0,1}∫ ln x * cos(t*ln x) / ln x dx = {0,1}∫ cos(t*ln x) dx After solving (I spent like 15 minutes and couldn't figure it out tbh, so I just used wolfram) you get: F'(t) = 1 / (t^2 + 1) We see that F(0) = {0,1}∫ sin(0) / ln x dx = {0,1}∫ 0 dx = 0. Therefore: I = F(1) = F(1) - F(0) = {0,1}∫ F'(t) dt = {0,1}∫ 1 / (t^2 + 1) dt = arctan(1) - arctan(0) = π/4

[@blackpenredpen]

It's 1:54am here. Good night!!!!!!!!!!!!

[@CarDealersdotcom]

Have a problem Mr D

[@yoavcarmel1245]

I solved it using I(t)=integral from 0 to 1 of sin(t*lnx)/lnx and got that right, maybe you could please upload a video using this method? If you would like, i can send you the picture of the solution somehow

[@blackpenredpen]

Flammy did that already... like 2 hrs after my upload, lolll

[@yoavcarmel1245]

blackpenredpen oh lol. Well done to him i guess :) will watch his video soon

[@yoavcarmel1245]

blackpenredpen no he did it different than me, i didnt use imaginary nums

[@snejpu2508]

U world is not powerful enough, but b world solves a problem. : ) YAY!

[@blackpenredpen]

Yup

[@hervesergegbeto3352]

Good morning sir How to become member ?

[@debrajbanerjee9276]

You can more easily do this by substituting ln(x)=-y which will leads to ...... I=∫(sin(y)e^(y))/y dy from 0 to ∞ now breaking sin(y) into taylore series and pulling the sigma notation out from the integral the integral will be a gamma function of (2n)! At last dividing it by (2n+1)! You will get series of arctan(u) with u=1 which immediately says that I=π/4

[@fengshengqin6993]

yeah ! right ! Good!

[@ianmoseley9910]

"more easily" - 😳

[@trace8617]

Ian moseley easier as in doesnt require complex analysis and identities such as ln(i)

[@BY-sh6gt]

Anyway how can u write integral sign in the comnent? 😂

[@FotisValasiadis]

ik its a bit too late folks,but i solved it in 5 minutes.Set u=lnx dx=e^u du its now (sinu*e^u)/u,use feynman's method to get rid of u by writing the integral as (sinu*e^(uy))/u and solve.You will end up with a simple sinu*e^(uy).Use the DI method and by the end of the day you end up with a -1 over (y^2+1) so you just know its an inverse tangent.you get that the original is minus inverse tangent plus π/4 so if you replace y=1 you get π/2-π/4=π/4.without any complex numbers having to step in

[@Jack_Callcott_AU]

Hey BPRP I really enjoyed that. It is very satisfying when complex maths leads to a simple result.

[@galgrunfeld9954]

Wow, that was so awesome! I haven't learned complex analysis, so I wouldn't think of expanding the scope to Complex numbers, that was clever! And when you zoomed in and I calculated the answer in my head, I was like "PFT, WHAT" and laughed, because the answer was so simple compared to how you solved it. One of the best videos of yours I've watched so far! :D

[@blackpenredpen]

Thank you!!! I am glad that you enjoy it!

[@ChefSalad]

You don't need complex analysis to learn about complex integration with exponentials. It's usually taught when doing differential equations. The reason is that it's way easier to do nonhomogeneous second order linear DE's using e^(ix) than with sin(x) and cos(x). To do ∫sin(x)dx, for example, you just do Im[∫e^(ix)dx] = Im[1/i*e(ix)] = Im[−i*e^(ix)] = −cos(x). It′s a bit overkill on a regular integral, but when doing nonhomogeneous second order DE′s, it′s a dream compared to the alternative method of undetermined coefficients. If you′re wondering what that looks like, I′ll give an example. Take x′′+2x′+x=sin(t). The characteristic equation is thus p(r)=r^2+2r+1=0, which means r=−1, twice. That′s makes the complementary solutions y₁=e^−t and y₂=t*e^−t. For the particular solution we can complexify the sin(t) as e^(it), thus α=i. We know that the particular solution has the form y*=e^(αt)/p(α), which means y*=e^(it)/(i^2+2i+1)=e^(it)/(2i)=−i/2*e^(it)=1/2*sin(t)−i/2*cos(t). Thus a particular solution is yₚ=−1/2*cos(t) and the whole solution is y=C₁e^−t+C₂te^−t−¹/₂cos(t). Finding the particular solution without using complex exponentials would involve solving a system of three equations or, even worse, a system of two equations with two integrals. This way just requires us to remember a simple rule.

[@leif1075]

Bit,can,younactually solve this without just knowing those formulas?

[@leif1075]

@Hassan Akhtar i wasnt being,salty..I,asked an intelligent question...to,see how to actually solve,this.why cant you see that..

[@mith_jain_here]

I was wondering the whole time how can an integral of a real function have a complex answer, but at the end when the answer simplified I was so relieved 😂. Maths is indeed beautiful.

[@axemenace6637]

This integral is very similar to 0 to infinity of sinx/x after the substitution x=e^u and the substitution I(a)=integral from -inf to 0 of e^au(sinu)/u. We want I(1). Feynman's technique solves this for us.

[@blackpenredpen]

Yea!

[@srpenguinbr]

First, I used u=ln(x) then used the feynman technique with I(t)= int from -inf to 0 of (sin(u)e^ut)/u

[@blackpenredpen]

Yup!!!

[@MarcLisevich]

Thanks!

[@blackpenredpen]

Thank you, too.

[@CornishMiner]

Some great techniques used to find a very satisfying answer. So good :)

[@blackpenredpen]

Yay!!

[@spudhead169]

ZOMG! That was a ride.

[@ayoubfenkouch5992]

this is why i like you videos , even if you understand the lesson very well you always surprise us with some tricks , but i have a question ( to you and to whoever reads this and can anwer me ) : when to think of such a method ? how to know if taking an integral to the complexe world and B world will give a results ? is there some hints within the integral ?

[@colinjava8447]

That's incredible, never seen that before, feynmann was a legend.

[@jschnei3]

The moment you plugged in b=-1 to solve for C, I slapped the table and shouted "You sneaky sonuvagun, you did it!!" That was an amazing moment

[@renesperb]

A different approach is to set t = ln x . Then you get the Integral of sin t/t*Exp(-t) ,( limits zero and inf.). Setting I[a]= sin t/t *Exp[-a*t] you can use Feynman's trick now to find the result π/4 .

[@ryanhurst5096]

Very creative problem solving process you used on this integral!

[@SVKODURU2008]

put ln x =-y, then, 0 to inf ∫e^-y siny /y dy= lim s=1 , s to inf ∫1/(s^2 +1) ds = π/2 -π/4 =π/4 , (Laplace)

[@The1RandomFool]

It's true that the limit as x approaches zero of x^a is undefined if the real part of a is zero. However, the limit becomes 0 if the real part of a is greater than zero. This can be proven by showing the limit as x approaches 0 of | x^a | is 0. Therefore, his step of 0^(1+bi) = 0 is valid.

[@weerman44]

Awesome integral! Thanks :D YAY

[@quidam3810]

Great video !!

[@premdeepkhatri1441]

Very very good solution of this problem

[@alanturingtesla]

Yay, I love these 20-minute integral videos!

[@blackpenredpen]

Yay!!!

[@AmanSingh-rg7hk]

Again tesla for turing.

[@alfredod.cadionjr.7035]

Video

[@silasrodrigues1446]

Oh my Gosh! This was really awesome! Brazilian congrats! #YAY

[@blackpenredpen]

Thanks!!!

[@varunsahni2128]

The general answer would be n*pi + pi/4 where n is integer. Which also state that area of this curve can take variable values

[@jorgesponja3042]

#YAY OMG I love how insane integrals ends with simple answers like pi/4 lol

[@blackpenredpen]

yay!

[@NurHadi-qf9kl]

Misal ln x=y maka dy=dx/x atau dx= x dy=e^ydy |=|e^y sin y dy= |sin y d(e^y) Lalu integral parsial.

[@mihaipuiu6231]

Beautiful solution!

[@TheBlueboyRuhan]

Good luck jaime for further maths

[@Tranbarsjuice]

Really cool integral and a very nice explanation

[@sirmac6726]

Aircraft trayectory: y = k / x k = 1 sqr km from: x1 = 0.5 km (y1 = 2 km) to: x2 = 2 km (y2 = 0.5 km) Velocity: V = const = 1000 km/h Max acceleration recommended a = 4 g a) Is the aircraft in danger? b) t=? time from x1 to x2.

[@skeletonrowdie1768]

I can only conclude it converges because at x=0 the integral is 0 (lnx>x for 0 to 1 domain). And finite everywhere else till x=1.

[@skeletonrowdie1768]

lol i already watched this video

[@jpradeesh3800]

If u know laplace transform, then proceed this way Put - ln(X) =t

[@abdullaalmosalami]

How does that help? Laplace transform I mean.

[@jpradeesh3800]

@@abdullaalmosalami you will an integral of form f(t) /t for which we have a formula. Then substitute s=1

[@Swybryd-Nation]

Euler evaluated this integral centuries ago by focusing on sin(ln(x)) first expanding it into an infinite series of sin(y) ie y-y^3/3!+y^5/5!-.......then you substitute y=ln(x)....ln(x) can be factored out and cancelled with the ln(x) in the denominator. Then it’s a simple ln(x) to a power evaluated term wise by Bernoulli first. Then you get the Leibniz series. Pi/4. Simples.

[@thatpersononline]

It's quite easy with Feynman's rule. I did it with g(t) = int sin(tlnx) dx/lnx evaluated at 0 to 1. Then evaluated g'(t) and complexified it. Pretty easy

[@ariusmaximilian8291]

Yay! This was so cool!! Thx for putting it up

[@-james-8343]

Hey awesome video, but you spelt the Jamie wrong in the title (you spelt it Jaime). Great video nevertheless, and keep it up!

[@blackpenredpen]

-James- thanks!! I just fixed.

[@-james-8343]

blackpenredpen no problem!

[@MrCuteguylol]

@@blackpenredpen jaime lannister?

[@yusufmia]

Hi. Your work is awesome

[@qbetech4764]

This can be also be done with F(a)= sin(alnx)/lnx with F(1)= I pretty easily. But complex world looks amazing.

[@omerangi4695]

That was very long and a very beautiful integral.

[@mohithalder3169]

8:55 "why don't we put a b here, and a b here, well u can try that but let me tell u it is enough" seriously ROFL after imagining.....😂😂😂😂

[@faresberarma3349]

Dear Blackpenredpen I'm one of your followers, and i noticed that you love complications lol, there's a more much easier way to do it look 1st let lnx=-t the integral become from 0 to infinity of sint*exp(-t)/t dt 2nd paramertrizing the integral by introduicing alpha in the sinus 3rd Derivate to alpha we obtain an easy integral (the laplacien of cos(at) The rest is a child game Bye

[@blackpenredpen]

Dear Fares BERARMA If you work out *all* the steps from your outlines, then it will take about the same amount of time/steps to *make it clear to the viewers* who haven't seen this kind of things before....... Bye

[@Linkedblade]

That was a wild ride from beginning to end

[@sy-py]

I solved it without complex number. I just used a Feynman's Trick to integrate sin(alnx)/lnx. For a=0 we get I(0)=0. I(1) is out integral in question. Now, I'(a)=integral of cos(alnx)dx from 0 to 1 which is 1/(a^2+1) (check for yourselves!)

[@kutuboxbayzan5967]

I think more easy way is I (b)=integral sin (blnx)/x 0 to 1 I'(b)=integral cos (blnx) 0 to 1 I' (b)=1/(1+b^2) I (b)=tan^-1 (b)+c And I (0)=0 =》c=0 I (b)=tan^-1 (b) I (1)=pi/4

[@sfarsi6]

Mathematician dressing code be like: for a supreme integral, I need a supreme shirt

[@wintersummers3085]

Math for its own sake is beautiful. Thanks blackpenredpen

[@blackpenredpen]

Winter Summers yay!!!

[@bijalshah9113]

I loved the way you solved this but I guess my method is easier... You can directly introduce a new variable: sin( 'b' lnx)/lnx, and then proceed with the same method. Finally you'll get: I(b) = arctan(b), where we want b=1, hence we get π/4. I hope that was helpful.

[@arolimarcellinus8541]

Why suddenly become arctan?? We don't know the definition of arctan though

[@srpenguinbr]

Wolframalpha told me 0^i is undefined. Can you do a video on that and maybe other complex limits?

[@howdoi_yt]

4:45 is there a way you can know where exactly to put the 2nd variable? or do you just keep trying to find the correct place?

[@SameerKumar-jf5mi]

this was fun! but in the end how do you know that log i yields π/2, and not something like 5π/2 ?

[@PunmasterSTP]

Supreme integral? More like "Super good video!" 👍

[@StefanDempf-x4s]

I don't understand how you can plug in -1 for b. Didn't you have earlier ...x^(i+bi) ? You did plug in x=0 and said the result was 0, but with b=-1 you have 0^0. Please explain. Love your videos

[@razielkeren6480]

why not uosing u substituting right away ? strat whit u=lnx and then the same method but no need for complex numbers. the Integrand will be (e^bu*sinu)/u

[@iOhadRubin]

That was actually pretty cool.

[@slahenejjari5334]

hy blackpenredpen this integral will be amazing in a vidieo: integral of (1/cos^n(x)) n natural

[@dimitris892000]

very good bprp, i suggest you try the integral from 0 to 2π of e^(cosx)* cos(sin(x)) dx #YAY

[@blackpenredpen]

Hmmm, I can try

[@codegurt5165]

You must've been out of breath by the end lmao

[@ikaros4425]

this is the kind of content I love to see, also why are you up so late???

[@holyshit922]

I(1) can be converted to arctan(1) This integral can also be calculated with Laplace transform Calculate L(sin(t)/t) and plug in s = 1

[@charlesnuett6674]

Hey blackpenredpen, what books or bdf books do you propose when one want to study integration ( I mean calculus in general) to the fullest just like you. So he can Know quite a lot. Pls 🙏🙏🙏

[@kepler4192]

*if you watched a video in the past* me who watched a video about feynman's technique 5 minutes ago: damn what kind of distant past is this

[@Hobbit183]

More multivariable calculus videos would be neat 🤙

[@wiwaxiasilver827]

There actually is another way in the final step. Using the definition that a+bi = re^(i*theta), with r being sqrt(a^2+b^2) and theta being arctan(b/a), or just the angle that forms on the Cartesian when the points are graphed with the x-axis are real and y-axis as imaginary, we get that ln(1+/-I) = ln(sqrt(2))+/-pi/4 (ln(r) + i*theta by logarithmic product rule and cancellation with e) but because it’s (1/(2i))(ln(1+I) - ln(1-i)), the parts with ln(sqrt(2)) cancel and we get pi/4 ultimately. Meanwhile, I guess this could be a cheat explanation but I think we can consider the 2pi*n of both thetas to mutually cancel in my way of calculation through subtraction.

[@Saki630]

What a wonderful first question on my exam.

[@Premkumar-nv8xf]

blackpenredpen becomes blackpenredpenbluepen

[@jasperh6618]

that was one heck of an adventure

[@blackpenredpen]

: )

[@gagadaddy8713]

Master Cao, u r way too clever ... Thanks a lot!

[@2thetutions153]

why we not using laplace properties L[f(x)/x]=integration of phi(x) from o to infinity.

[@hendrixgryspeerdt2085]

How do you know to take the principal value of ln(i)? Since there are infinite possible values for ln(i). i(Pi/2 + 2(Pi)n), n is an integer.

[@airatvaliullin8420]

I solved it by changing the variables: t = -ln(x). I got the integral from 0 to +inf of sin(t)/t*e^(-t). Then I used the series definition of sin(t) and swapped integration and summation (bcs I can :)). The improper integral was equal to (2n)! which I very liked. In the end, the sum was exactly the arctan(1) by the series definition. The steps (in latex code - feel free to paste into desmos for readability): integral = \int_{0}^{+\infty}\frac{1}{t}\sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{\left(2n+1 ight)!}t^{2n+1}e^{-t}dt = \sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{\left(2n+1 ight)!}\int_{0}^{+\infty}t^{2n}e^{-t}dt = \sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{2n+1} = \arctan1 = \pi/4

[@fengcheng3507]

Yes, this can be converted into the Laplace transform of (sin(t)/t), and the result = π/2 - arctan(S) with S=1, i.e. π/4.

[@VaradMahashabde]

IS THIS RIGHT? Using your complex definition of a sin, I took the e^i common and rewrote i as e^(iπ/2), obtaining sin z = (e^ln z - e^ -ln z)/2e^(π/2) If z = ln x, We get sin ln x = x/e^(π/2) _(Which is like that shouldn't be right ???????)_

[@Samir-zb3xk]

This can actually be solved without complex numbers Place the parameter in the argument of sin; then differentiate You'll then have to solve the integral of cos(t•ln(x)), which looks pretty intimidating but it can be solved with a u-sub and then integration by parts You then get I'(t)=1/(1+t²) I(t)=arctan(t)+c; but I(0)=0 so I(t)=arctan(t) I(1)=π/4

[@EAtheatreguy]

I did this integral without bringing in complex numbers at all, quick u-sub u=lnx turns it into int(e^(u)sin(u)/u)du from negative infinity to 0, parametrize with I(b) = (e^(bu)sin(u)/u)du from negative infinity to 0, proceed. I got the wrong answer a few times because when solving for the constant at the end, I let b go to negative infinity instead of positive infinity, which is wrong because u is always negative, so letting b go to negative infinity actually causes divergence.

[@kmac5912]

May you please make a video on how to solve for x=y^2+x^2y y=x^2+y^2x

[@Nickesponja]

But isn't ln(i)=iπ/2+2kπ for integer k? But this integral clearly cannot have more than one answer. Am I missing something?

[@Darkev77]

Let me know if you the answer

[@crysiswar7632]

Because x is between 0 and 1

[@6612770]

Wow, but Phew! I'm exhausted after watching that marathon.

[@blackpenredpen]

: )

[@dimosthenisvallis3555]

That was pretty awesome. Hey, could you tackle this one--> 4(x^2) + x + 1 = ((2x - 1)^(1/2))((x+1)^(1/4)). i was able to prove there r no real solutions. but what about complex solutions. I think this could be an intresting video. Keep up the awesomeness

[@sareprne]

Is it wrong for me to take parameter b for both the x variables in the numerator of I(b) such as _x^(ib)-x^(-ib)_ ?

[@williamliamsmith4923]

Refer to video about lim(x->0) {sin(ln x)/ln x} = DNE. Can we observe the convergence of this integral to determine lim(x->0) {sin(ln x)/ln x}? I am thinking: If the area under {sin(ln x)/ln x} is finite (for 0

[@kunalbatra4166]

loved this one..

[@jimnewton4534]

One thing that I don't understand about this derivation is: after you introduced b, you had x^(bi)-x^i in the numerator. Then you took the derivative and the x^i went away. HOWEVER any second term not involving b would have also gone away. So if you had started with any x^(bi) - f(x) would you have still gotten the same answer? Is that troublesome?

[@jimnewton4534]

I suppose the answer is hidden somewhere in finding the "constant" C, which is not really a constant, but rather is a function of x, right? Perhaps that is some subtlety which needs to be discussed, that C is not a constant, but is only constant with respect to b?

[@nordgothica]

Where does 3^t.ln(3) come from? Don't you just have the exponent times whatever is in the power differentiated? So you'd have 3^t.(1) = 3^t.

[@Γιώργος-β2τ]

This is so perfect

[@alegian7934]

Really cool integral. Can someone please explain why we use lni=πi/2 and not any of the other values?

[@alexanderpanov2326]

Ln e^i•п/2=п/2•i•lne= п/2•i

[@HoldensBro]

So I did feynman’s trick for ∫(sin(alnx)/lnx)dx Then f’(a)= ∫ cos(alnx)dx Then I did a u-sub for u=alnx Then I got (1/a) ∫ cos(u)(e^(u/a))du with limits at -infinity and 0. Then I evaluated the integral using partial integration (DI method). Eventually this led to (sin(u)(e^(u/a))a^2+cos(u)(e^(u/a))a)/a(a^2+1) Evaluating at 0 gives 1/(a^2+1) and evaluating at -infinity gives 0. Then integrating with respect to a gives arctan(a)+c. To find c, sub a=0 into the original integral, giving 0. Since arctan(0)=0, c must also be 0. Thus, the integral is arctan(1), which is pi/4. I don’t have any clue whether this was more efficient, but it worked.

[@TonyStark-30001]

Hello blackpenredpen plz solve the MIT intgration bee Question plz sir

[@TheMiningProbe]

This was an extremely clever method, you have my applause

[@blackpenredpen]

Yay!!! Thanks to Jamie tho! : )

[@srikanthtupurani6316]

very nice. this is a nice technique called complexification. but you should explain why we can take the derivative inside the integral sign. math subject is like that. we need to justify everything.

[@lindsaywaterman2010]

This would have been much simpler if you made ln(x) = -v and then use Feynman's technique! I(a) = Integral of [sin(v)/v]*e^-av between 0 and infinity.

[@mathunt1130]

Why can't you do the obvious of making u=-log(x), to get the integral of exp(-u)sin(u)/u from 0 to infinity? Then consider the integrand exp(-au)sin(u)/u, and then differentiate w.r.t a to get an integrand of -exp(-u)sin(u), which can be easily solved using repeated integration by parts.

[@manavshah1084]

this can be done more easily by using feynmen's trick in the first step itself taking I(b)=integration(sin(blnx)/lnx)

[@manavshah1084]

using that we won't have to deal with complex numbers, they will cancel out and just give arctan(b) as I(b)

[@gabrielenzian6475]

I think it would be better to use the archangent and not complexes just for simplicity and speed, thanks.

[@shanmugasundaram9688]

The convergence and continuity of the function sin(ln x)/ln x at x=0 need to be discussed.

[@abhishekmewar671]

We can take " lnx=t"and solve the equation using Laplace transform occupying only half a page. That said love your channel keep up the good work.

[@alxanderjon8716]

How would you do that? Doesn't Laplace need initial conditions? Ive only seen it in the context of differential equations.

[@MG-hi9sh]

@@alxanderjon8716 The definition of the Laplace transform is actually in the form of an integral. If you use that definition, it could work. That's my thinking anyways.

[@MG-hi9sh]

@@alxanderjon8716 Laplace can be used for integrals as well, it isn't just a DE method.

[@alxanderjon8716]

@@MG-hi9sh ok so if we do the substitution and change the bounds we have integral from -inf to 0 of e^t(sint) then if we take the laplace transform of this we will have a double integral, but both wrt t. How does that work? do you just combine the limits?

[@MG-hi9sh]

@@alxanderjon8716 No, Laplace would change the variable to s and then you would change bounds based on s values. Laplace of e^tsin(t) is 1/((s-1)^2)+1), and you integrate using that, I think. Don't take my word for it though, but I think that would be the reasoning.

[@Mathmagician73]

Great sir 😁😁😁😁👌👌👌👌😳😳u r doing very great work but i have another method to solve this integral can i share u?

[@kshitijarora382]

Please solve for x in x= tan^-1(i)

[@MrBoubource]

That supreme jacket tho