Supreme Integral with Feynman's Trick

[@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?

[@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

[@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

[@snejpu2508]

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

[@blackpenredpen]

Yup

[@hervesergegbeto3352]

Good morning sir How to become member ?

[@Jack_Callcott_AU]

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

[@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!!!

[@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..

[@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!

[@CornishMiner]

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

[@blackpenredpen]

Yay!!

[@MarcLisevich]

Thanks!

[@blackpenredpen]

Thank you, too.

[@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.

[@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

[@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 .

[@alanturingtesla]

Yay, I love these 20-minute integral videos!

[@blackpenredpen]

Yay!!!

[@AmanSingh-rg7hk]

Again tesla for turing.

[@alfredod.cadionjr.7035]

Video

[@colinjava8447]

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

[@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 ?

[@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.

[@ryanhurst5096]

Very creative problem solving process you used on this integral!

[@wintersummers3085]

Math for its own sake is beautiful. Thanks blackpenredpen

[@blackpenredpen]

Winter Summers yay!!!

[@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

[@weerman44]

Awesome integral! Thanks :D YAY

[@jorgesponja3042]

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

[@blackpenredpen]

yay!

[@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)

[@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

[@TheBlueboyRuhan]

Good luck jaime for further maths

[@Tranbarsjuice]

Really cool integral and a very nice explanation

[@6612770]

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

[@blackpenredpen]

: )

[@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.

[@omerangi4695]

That was very long and a very beautiful integral.

[@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

[@quidam3810]

Great video !!

[@premdeepkhatri1441]

Very very good solution of this problem

[@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

[@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.

[@TheMiningProbe]

This was an extremely clever method, you have my applause

[@blackpenredpen]

Yay!!! Thanks to Jamie tho! : )

[@qbetech4764]

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

[@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.

[@ariusmaximilian8291]

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

[@spudhead169]

ZOMG! That was a ride.

[@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 ?

[@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?

[@mihaipuiu6231]

Beautiful solution!

[@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

[@Linkedblade]

That was a wild ride from beginning to end

[@2thetutions153]

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

[@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

[@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.

[@jasperh6618]

that was one heck of an adventure

[@blackpenredpen]

: )

[@-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?

[@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.

[@EMorgensztern]

can you prove the continuity of y=x^(1/x) pls ?

[@Humongastone]

At 14:02 , he said something about negative?? What is that about?

[@blackpenredpen]

Oh bc usually when we integral 1/(1+x) we get ln|1+x|, But I said since the ln inside had complex number, so dont worry about the abs value. : )

[@Humongastone]

Oh ok thanks!

[@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.

[@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?

[@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.

[@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!)

[@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

[@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?

[@anuj-ios]

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)_ ?

[@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 🙏🙏🙏

[@kmac5912]

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

[@sfarsi6]

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

[@Saki630]

What a wonderful first question on my exam.

[@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

[@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.

[@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.

[@Quwertyn007]

...can someone explain to me why ln(i) isn't πi(1/2+2k)? ^^'

[@abhisarma7249]

It is, because ln is a periodic function in the complex world, but in this context of real integration it only makes sense to take the principal value

[@Quwertyn007]

Abhi Sarma Thanks for the response, it did clear up some of my confusion. ^^ And now I see that in a sense these infinitely many solutions are caused by the fact that the ln in the integral on the very beginning could be interpreted this way as well. But I still don't get how do we know it's this value for ln(i) and not a different one...

[@Quwertyn007]

Hamish Blair But the integral itself only has one answer (assuming we mean the real logarithm of x)

[@emmeeemm]

I think the way I would describe this, there are countably infinitely many analytic solutions, but only one solution consistent with other concepts, such as graphing this real-valued function and applying the concept of Riemann sums that generates the Riemann integral. Sure, in a vacuum, you might want to account for the whole family of solutions. But when the domain is restricted to the Real numbers, the function y(x)=sin(ln(x))/ln(x) has a unique Cartesian graph, and we can use other concepts of mathematics to narrow down what "the" solution to this integral is, which we would get if we had a method for it that did not leave the realm of the Real numbers. I've prepared a demo of the Riemann sum here: www.desmos.com/calculator/p5mh9so1hl Just move the slider for k to watch the value of the Riemann sum change. And compare it to the integral also calculated by Desmos.

[@angelmendez-rivera351]

Do the integral of e^x*sin(ln x) from x=0 to infinity ?

[@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

[@PunmasterSTP]

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

[@yusufmia]

Hi. Your work is awesome

[@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 ???????)_

[@azmath2059]

Sensational

[@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.....😂😂😂😂

[@rowechenzhong8950]

Wait a minute. I'm pretty late, but it's much faster and cleaner this way. You end up with (1/2)(1/(1-ib)+1/(1-ib))=1/(1+b^2), which results in arctan(b). Then arctan(1)=pi/4 done.

[@kunalbatra4166]

loved this one..

[@user-nb6zu3rk4f]

19:10 How did WolframAlpha find the integral?

[@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

[@ikaros4425]

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

[@attepiltonen6607]

Nice

[@arequina]

I remember this integral back in college several decades ago...indef int of [ ln (x^2)/(1+x^2) ] dx....never got the correct answer. Would love to see what this is.

[@azmath2059]

Just plug it into the integral calculator www.integral-calculator.com and show steps

[@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

[@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.

[@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

[@shanmugasundaram9688]

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

[@GreenMeansGOF]

Sooooo.... who knows the restriction for b? I understand series convergence but I dont think im familiar with integral convergence.

[@Hobbit183]

More multivariable calculus videos would be neat 🤙

[@slahenejjari5334]

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

[@theimmux3034]

A non-complex, perhaps less exciting approach: Begin with u = lnx u = lnx x = e^u dx = e^u du Now we have: ∫ (sinu / u * e^u) du. Let's define a function I(a) such that I(a) = ∫ (sinu / u * e^(au)) du. Notice that I(1) is equal to the original definite integral at hand. Let's differentiate both sides with respect to a: I(a) = ∫ (sinu / u * e^(au)) du I'(a) = ∫ (sinu / u * ue^(au)) du I'(a) = ∫ (sinu * e^(au)) du Perform integration by parts so that the integrand repeats. This yields the following equation: (1 + a^2)∫ (sinu / u * e^(au)) du = -1 ∫ (sinu / u * e^(au)) du = -1/(1 + a^2) I'(a) = -1/(1 + a^2) Time to integrate both sides with respect to a. It is a well known result that ∫ 1/(1 + x^2) dx = tan^(-1)(x) + C. Let's use that piece of information to our advantage: ∫ I'(a) da = ∫ (-1/(1 + a^2)) da I(a) = -tan^(-1)(a) + C Recall that I(a) is defined as I(a) = ∫ (sinu / u * e^(au)) du. If we plug in a = 0, we get the definite integral of sinu / u from negative infinity to zero which is essentially the same thing as going from zero to positive infinity. This is because sinx/x is an even function. The value of this definite integral is known very well to be equal to π/2. We could also let a go to negative infinity and carry on from there but things get a little akwards that way. I believe that also works but it is much simpler this way. Let's continue with a = 0: I(0) = -tan^(-1)(0) + C π/2 = C C = π/2 Now that we know another representation for I(a) besides it's original definition, let us finally plug in a = 1. If you remember, this is equal to the definite integral going from 0 to 1 of sin(lnx)/lnx. That's what we are trying to solve. I(1) = -tan^(-1)(1) + π/2 = -π/4 + π/2 = π/4 In conclusion, I(1) = ∫ (sin(lnx) / lnx) dx (from 0 to 1) = π/4.

[@richtw]

Awesome!

[@ethanchandler3934]

If you use other values of ln(i), what would this mean intuitively? I believe if we add 2pi, it ends up being complex. What do complex values of integrals mean? Is this the area of the function but on a different slice of the plane?

[@davidepeccioli4431]

I may be wrong, but I assume that the ln function still refers to the inverse of the e^x function, knowing that e^(iπ)+1=0 (Euler identity)

[@TonyStark-30001]

Hello blackpenredpen plz solve the MIT intgration bee Question plz sir

[@Γιώργος-β2τ]

This is so perfect

[@dolevgo8535]

Maybe its too late. But how can i know i can choose ln(i) to be i*pi/2? I mean, the integral should have only one value but we could choose 5*i*pi/2 and so on..

[@gabrielenzian6475]

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

[@joshuaiosevich3727]

A much easier solution is to set u equal to ln(x), then set y=-u, then you get the integral from 0 to infinity of e^-xsin(x)/x and you realize we're done because this just requires the same feynman trick that gets us the solution to the dirichlet integral. define I(t)=e^-xt*sin(x)/x, I'(t)=-1/(1+t^2), I(t)=-arctan(x)+C, I(infty)=0, thus C=pi/2, plug in t=1 and pi/2-arctan(1)=pi/2-pi/4=pi/4, and ya done.