Complex Integration: The ML Inequality Proof and Example

This video proves the ML Inequality (aka Estimation Lemma) for complex integrals and does a short example involving it. The ML Inequality is quite useful because it helps establish upper bounds on your complex/contour integrals. It was a request that someone made yesterday, so the fact that I managed to complete the request in one day is probably quite fortunate lol.

Questions/Requests/Feedback? Let me know in the comments! Also, I got a new microphone, which is why my voice sounds much better now.

[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!

[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!

[sarthakgaur]

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.

[catprincess]

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!

[sagarawasthi]

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)

[ozzyfromspace]

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

[Geneve]

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 :)

[pip]

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

[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.

[CludEater]

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.

[amandeep]

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

[leonig]

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.

[laurencereeves]

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..

[hamzaehsankhan]

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

[jaimecaballero]

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

[ameepatel]

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