Complex Integration: The ML Inequality Proof and Example

[@FacultyofKhan]

Just realized a mistake I made at 10:25. The actual value of the contour integral is, in fact, pi/2*i (so the magnitude is pi/2), which still obeys the ML inequality theorem. You could verify this by using the Residue Theorem and integrating over the whole circle. Since the residue of 1/z is just 1, the residue theorem would give the contour integral around the entire circle as 2*pi*1*i. One-quarter of that (this arc C in the example), is pi/2*i, which is consistent with the actual answer. Nevertheless, the ML inequality is NOT to be used as an approximation, even though in this case it gave the actual answer. My apologies for the dumb mistake, but I'm hoping it won't ruin the video since it's just a small footnote at the end!

[@susiehue9465]

can't it also just be solved saying that the integral of 1/z is ln(z) and evaluating it from 2i to 2? Don't we then get ln(i) which is i pi/2?

[@avdrago7170]

Susie Hue no that doesn’t always works, it may have just worked this time

[@klnrdknt]

Dude, your lectures make a whole semester of math in an understandable way at maybe 5% of the normal time. I wish my profs would watch some of your videos for inspiration. Thank you!

[@sarthakgaur8018]

To the point and very helpful video. Good that you didn't make it too long considering that the university students do not have much time for it.

[@catprincess9]

Wow that was a really quick fulfillment of my request! Thank you so much! My doubt is regards using the residue theorem in solving integrals on a line which extends from -R to +R and R tends to infinity. Then if you integrate around a semicircular loop you usually can show the integral along the semicircular arc of radius R is 0 as R tends to infinity by using the ML inequality. To show this, there is a theorem that states the degree of denominator should be at least 1 larger than the degree of the numerator. But if that was the case when you use ML inequality to upper bound your integral, it would introduce one more R in the numerator as the length of the semicircular arc is pi*R, and now your numerator and denominator would have the same degree and therefore the integral would not be 0 as R tends to infinity. (Michael Barrus used this to solve real integrals (2/2))Thanks!

[@FacultyofKhan]

Glad you liked it! I'm not sure if you're asking a question there, in which case you may have to be more specific since I don't know what (denominator degree) theorem you're referring to. In any case, I hope that this video helped with your research. Thank you for the feedback!

[@catprincess9]

I am sorry I did not explain myself very well. My question was with regards to using ML inequality to prove that the integral of a (certain kind of) function on a semicircular arc is 0 as the radius of the arc tends to infinity. One usually encounters such situations while using residue theorem to solve real integrals. The name of the theorem I was referring to was not mentioned where I cam across it. Nevertheless, thank you very much for responding to my (oh so many!) comments. And good luck with your thesis defense!

[@FacultyofKhan]

Ah I see. I don't know if I'll be able to address your problem directly without knowing what the function is, but I hope that my videos helped in getting you to your end goal! And thank you for the kind wishes; I appreciate it!

[@catprincess9]

Just found out. It is called Jordan's lemma. I was approaching it the wrong way. Thanks anyways!

[@PhysicsMath]

Epsilon delta please..

[@pip119]

At 2:42, is there any particular reason Khan takes the exponential term inside the integral? (As a constant, this requires no justification?).

[@FacultyofKhan]

Because the exponential term is a complex number and I wanted to move the Re inside the integral. I can't really do that if there's a complex number sitting outside.

[@harshitgupta3386]

@@FacultyofKhan It's an incorrect step as theta is a function of t, so you can't take it inside the integral. You can however, put the inequality there and continue with the proof in the same manner as it is shown in the video. Both of them will yield the same result

[@kosalaherath1350]

@@harshitgupta3386 No, theta is not a function of 't'. Since r*exp(theta) equals to the integration over t, theta and r have absolute values (constants for the given path). Therefore, we can take exponential term inside the integral for that given contour path. In other words, theta value does not vary with the 't' value, it got from all the available t values over [a,b].

[@Geneve141]

Thank you for your beautiful and helpful work dear Khan. I have trouble in applying the residue theorem to solve a real integral like this one : "integral of cos(ax)/(1+x^2) dx from 0 to infinity" where a is a positive real number. Maybe you find some time to explain us this application :)

[@FacultyofKhan]

Sure thing! I will make a video about applying the residue theorem once I get the time! Thank you for the kind feedback!

[@ozzyfromspace]

I feel like this theorem is intuitively obvious, nevertheless, it’s super useful so I yield

[@laurencereeves8626]

at 2:12 you write that the magnitude of e^(itheta) is the equal to the magnitude of cos(x)+isin(x), however this would be equal to the sqrt of (cosx)^2 - (sinx)^2 no?? If you take i^2 it becomes a negative. Hence, I am a little confused how you set the magnitude of e^itheta to 1..

[@shook1364]

wouldn't it be √(cos²x + sin²x)? |z| = √(Re(z)² + Im(z)²) Re(z) = cosx Im(z) = sinx

[@amandeep9930]

if we take f(z)=1 and the contour gamma(t)=iota*t where t is in [0,1] then we see that we cannot take Re inside the integral...can someone please explain why this does not contradict the proof

[@sagarawasthi3631]

I highly appreciate your efforts, I have one request that if you can make problem solving vides also (at least one problem(good) from each topic)

[@hamzaehsankhan]

Isn't there a difference between polar form and euler form?

[@FacultyofKhan]

The polar form of a complex number (z = r*exp(i*theta)) can be directly applied to the Euler formula so that z = r*cos(theta) + i*r*sin(theta). I haven't heard of the term 'Euler form' but you can certainly use the Euler *formula* on the polar form quite readily.

[@alicejbbennett]

Great video! You should look at getting a pop filter for your microphone though, the P sounds come out a bit in the audio.

[@FacultyofKhan]

Thank you for the feedback! I'll think about getting a pop filter, but I'm wondering if there's a way in Audacity to get rid of the pop without having to buy a filter. An Audacity trick would probably be more practical.

[@FacultyofKhan]

I hope you didn't find the sound too bad though (at the very least, better than the sound from my other videos lol).

[@alicejbbennett]

I wouldn't know about audio editing, but it's worth looking into.

[@Cl0udEater]

I don't know what the "ML" actually stands for, but I can't help but think "Marxism-Leninism" inequality in my head. In any case, this video was helpful.

[@j_blue6784]

M and L are variables (M is the maximum of |f| on the path and L is the length of the path)

[@Cl0udEater]

@@j_blue6784 ahh, thanks! That was back in the Autumn quarter, but glad to finally know that. : ]

[@ameepatel6898]

hello sir mujse ek example solve nai hota please help me Sir

[@jaimecaballero8267]

My only doubt is That integral should not have a circle in the center?

[@FacultyofKhan]

I skimmed the video and didn't see an integral with a circle in the center (which refers to an integral over a closed curve). Can you give a timestamp of the part you're referring to?

[@j_blue6784]

@@FacultyofKhan I assume that he/she thinks that there is a circle missing (which is not the case as far as i know)

[@leonig100]

Enjoy your videos immensely. I have a query, At 2.00 you set integral w(t)dt to r0e^iTheta0. What is the justification for this? After all w(t) can be any complex number.

[@jorgeestebanmendozaortiz873]

If you remember that e^iTheta0 = cosTheta0 + i sinTheta0, then r0e^iTheta0 is just the polar representation of the given integral, where r0 is the magnitude of the resulting complex number and Theta0 is the angle of such number.

[@lateefahmadwanilaw8948]

Very much Interesting

[@FacultyofKhan]

Glad you enjoyed it!

[@scholar-mj3om]

Marvelous💯💯

[@MegaBdboy]

What is contour ?

[@FacultyofKhan]

It's a curve on the plane basically.

[@rashisardana1853]

Really nice

[@abhirajeevam1672]

Thank you😘😘😘

[@MBAzeBest]

Hoopla

[@gallergaller9889]

This is too hard