Lecture 12 | Integration through a Branch Cut | Example 1 | Theta Classes

From this lecture we are going to learn about how to solve an improper integral involving multi valued function. So firstly we will discuss about branch point, branch cut in a brief manner and then we will solve given integral. While solving integrals involving multi valued function we will learn how to avoid branch point when it comes on the boundary of the contour and we are also going to see how to select a branch cut. After that we will find residue and using Cauchy's Residue Theorem we will solve the integral.
In case if anyone has any doubt regarding this lecture do mention it in the comment section.
Keep connected with Theta Classes to study various topics of higher mathematics.

[kunjalvara]

Your way of teaching is excellent. It is vary difficult part of complex integration, and your explanation is just mind blowing. Thank you very much sir.

[Legacies]

Thanks, It helped a lot. It was lucid explanation and now all doubts are clear.

[hiralaldebnath]

Sir you are a living legend... thank you so much....🤩

[apnaphysics_]

Thanks sir for simplyfying the most confusing part of complex integration, I enjoyed the whole lecture

[subhasmrutipanda]

Sir syllabus me likha hua he ki contour integration involving branch point toh konsa aur kese karna he please samjha dijiye

[piyalikarmakar]

please make a video on 'infinite products and theorems on it' e.g. product (1+an) converges iff sum[log(1+an)] converges

[jyotijangra]

Sir 34:42 pr hmne z=1/t substitute kyu kiya h, agr hm sidha z ki limits put krte h tb b to 0 hi aa rha h

[ShubhamSharma-hxdf]

30:00 ka agar use hi nhi karna hai to uska likhna zaruri hai kya?
Jo Ln π + 0i hai vo vala

[holyshit]

This integral can be calculated without complex analysis
Int(ln(x)/(x^2+1)^2, x=0..infinity)
u=1/x
du=-1/x^2dx
du=-u^2dx
dx = -du/u^2
Int(ln(1/u)/(1+1/u^2)^2(-1/u^2), u=infinity..0)
Int(-ln(u)/(1+1/u^2)^2(1/u^2), u=0..infinity)
Int(-ln(u)/((u^2+1)/u^4)(1/u^2), u=0..infinity)
Int(-ln(u)u^4/(u^2+1)^2(1/u^2), u=0..infinity)
Int(-ln(u)u^2/(u^2+1)^2, u=0..infinity)
Integration by parts
1/2*uln(u)/(u^2+1)|_{0}^{infinity}-1/2Int((ln(u)+1)/(u^2+1), u=0..infinity)
-1/2Int(ln(u)/(u^2+1), u=0..infinity)-1/2Int(1/(u^2+1), 0..infinity)
1/2(Int(ln(u)/(u^2+1), u=0..infinity)+Int(ln(u)/(u^2+1), u=0..infinity))
1/2(Int(ln(u)/(u^2+1), u=0..infinity)+Int(ln(1/u)/(1/u^2+1)(1/u^2), u=0..infinity))
1/2(Int(ln(u)/(u^2+1), u=0..infinity)-Int(ln(u)/(1+u^2), u=0..infinity))
So we have integral
-1/2Int(1/(u^2+1), u=0..infinity) = -π/4

[RohitYadav-ivze]

Bhaiya aab mere liye bhagwan ki tarah hai kya padhate ho guru aap . Bhaiya plzz vector intigration ka Video dal dijiye