maths 4

1. Application of Residue Theorem to
evaluate real integrations. Name of Guide:
Prof. Ankita Mane
NAME ROLL NO
Akansha Mahadev Gore
Shweta Satish Desai
Sahil Deepak Shah
Yashoda Keshav Poojari
54
31
59
18

2. Outline
• Introduction
• Concept
• Problem
• Conclusion
• Reference

3. Augustin Louis (Baron de)
Cauchy

4. Introduction
Augustin Louis (Baron de) Cauchy was a French mathematician of great repute who
contributed to various branches of mathematics. He wanted to be an engineer, but because
of poor health, he was advised to pursue mathematics. His mathematical work began in
1811 when he gave brilliant solutions to some difficult problems of that time in the next 35
years, he published 700 papers in various branches of mathematics. He is supposed to have
initiated the era of modern analysis.
• Cauchy’s Residue Theorem
If f(z) is analytic inside and on a simple closed curve C,
except at a finite number of isolated singular points z1,
z2 ….. zn inside C then,
C
f z ⅆz = 2πⅈ [sum of resⅈⅆue at z1, z2 …. zn]

5. Concept
Type I: Integral of the form ∫129_(−∞)^∞𝐟(𝐱)ⅆ𝐱=∫129_(−∞)^∞〖𝐏(𝐱)/𝐐(𝐱) ⅆ𝐱〗 where P(x)
and Q(x) are polynomials and the degree of P(x) is greater than
the degree of Q(x) at least by 2.
Consider the integral ∫129_cP(z)/Q(z) dz where C is the closed contour consisting of a
semi – circle C_1 of radius R in the upper half with center at the origin and the part
of the real axis from – R to + R.
We assume that there is no singularity on the real axis and we take R sufficiently large
so that all the singularities of f(z) lying in the upper half of the z – plane lie within the semi
– circle.
Now, by Cauchy’s Residue Theorem,
c
f z ⅆz =
c
P z
Q z
ⅆz =
c1
P z
Q z
ⅆz +
−R
R
P z
Q z
ⅆz
= 2 π ⅈ [sum of the residue at poles in the semi – circle] …. (1)

6. Now, as R = ∞ i.e., z = ∞ if z f(z) = 0 i.e., z
P z
Q z
= 0 then we have by Cauchy’s Lemma
c1
P z
Q z
ⅆz = c1
f z ⅆz = 0 in the limit.
Hence, from (1) we get, as R = ∞.
c
f z ⅆz =
−∞
∞
P z
Q z
ⅆz =
−∞
∞
P x
Q x
dx = −∞
∞
f(x) ⅆx
∴ −∞
∞
f(x) ⅆx = 2 π i [Sum of the residues at poles of f(x) lying in the upper half]

7. Type II: Integral of the type
−∞
∞
𝐜𝐨𝐬 𝐦𝐱
𝛟 𝐱
𝐝𝐱 𝐨𝐫
−∞
∞
𝐬𝐢𝐧 𝐦𝐱
𝛟 𝐱
𝐝𝐱 𝐰𝐡𝐞𝐫𝐞 𝛟 (x) is a polynomial in x.
We start with c
F z ⅆz = c
eimz ⋅ f z dz where f (z) =
1
ϕ z
. As in the previous case we consider a contour
C consisting of a semi – circle C1 of radius R with centre at the origin lying in the upper half of the z –
plane and the part of the real axis from – R to R. We assume that there is no singularity on the real axis and
we take R sufficiently large, so that all the singularities of f (z) lying in the upper half of the z – plane lie
within the semi – circle.
Now, by Cauchy’s Residue Theorem,
c
f z ⅆz = c1
f z ⅆz + −R
R
f(z) ⅆz
= 2 𝜋 i [Sum of the residue at poles in the semi – circle]
Now, as R = ∞ i.e., z = ∞, if f (z) = 0, we have by Jordan’s Lemma c1
f z ⅆz = 0 in the limit.
∴ 𝑐
𝐹 𝑧 ⅆ𝑧 = −∞
∞
𝐹 𝑧 ⅆ𝑧 = −∞
∞
𝐹 𝑥 ⅆ𝑥 = −∞
∞
ⅇⅈ𝑚𝑥
𝐹 𝑥 ⅆ𝑥 = 2 𝜋 i [Sum of residues]
∴ −∞
∞
ⅇⅈ𝑚𝑥
𝐹 𝑥 ⅆ𝑥 = 2 𝜋 i [Sum of residues]

8. Thank you