Lecture Notes of Mathematics-I
for Integral Calculus, Improper
Integrals, Beta and Gamma
functions
by
Prof Shiv Datt Kumar
Department of Mathematics
Motilal Nehru National Institute of Technology
Allahabad (UP), India
Pin - 211004
E-mail: sdt@mnnit.ac.in
ii
Contents
1 Integral Calculus
1.1
v
Jacobian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1
Change of variables
v
. . . . . . . . . . . . . . . . . . . . . . vi
1.2
Volume of solid of revolution . . . . . . . . . . . . . . . . . . . . . viii
1.3
Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
1.3.1
1.4
Change of variables in double and triple integral . . . . . . xi
Triple Integral
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . xii
2 Improper Integrals, Beta and Gamma Functions
xvii
2.1
Improper Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvii
2.2
Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xx
2.2.1
2.3
Gamma Functions . . . . . . . . . . . . . . . . . . . . . . . xxi
Dirichlet Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxiii
Bibliography
xxv
iii
Shiv Datt Kumar
Integral Calculus
iv
Chapter 1
Integral Calculus
1.1
Jacobian Matrix
Let F : Rn −→ Rm be a vector valued function given by F (x1 , . . . , xn ) =
(F1 (x1 , . . . , xn ), . . . , Fm (x1 , . . . , xn )). Then Jacobian matrix is the matrix of all
first order partial derivatives defined as

∂F1
∂x1
..
.



J(F ) = 

 ∂Fm
∂x1
...
..
.
...
∂F1
∂xn
..
.
∂Fm
∂xn







Jacobian matrix is important because if the function F is differentiable at point
p = (x1 , . . . , xn ), the Jacobian matrix defines a linear map Rn −→ Rm , which
is the best linear approximation of the function F near point p. The Jacobian
generalizes the gradient of a scalar valued function of several variables, which is
generalization of derivative of a scalar valued function of single variable. Jacobian
can be thought of as describing the amount of stretching, rotating or transforming
and that transformation imposes locally.
Definition 1.1.1. If F1 , F2 , . . . , Fn are functions of x1 , x2 , . . . , xn , then the dev
Shiv Datt Kumar
Integral Calculus
terminant
∂F1
∂x1
.
J = ..
...
..
.
∂Fn
∂x1
...
∂F1
∂xn
..
.
∂Fn
∂xn
is called Jacobian of F1 , F2 , . . . , Fn with respect to x1 , x2 , . . . , xn , In short it is
∂(F1 , . . . , Fn )
written as
.
∂(x1 , . . . , xn )
Remark: Determinant of square Jacobian matrix (called as Jacobian) gives
important information about the behaviour of F near that point. If the Jacobian
determinant at p is non-zero, then the continuously differentiable function F is
invertible near a point p ∈ Rn . This is the inverse function theorem. Further if
Det(J) > 0, then F preserves orientation near p. If Det(J) < 0, then F reverses
orientation. Absolute value of the Jacobian determinant gives the factor by which
the function F expands or shrinks volumes near p.
Properties of Jacobian:
1. If f and g are functions of u and v and u, v are functions of x and y, then
∂(f, g)
∂(f, g) ∂(u, v)
=
∂(x, y)
∂(u, v) ∂(x, y)
′
2. If J is the Jacobian of the system u, v with respect to x, y and J is the
′
Jacobian of x, y with respect to u and v, then JJ = 1.
Example 1.1.2. If x = rcos θ and y = r sin θ, then (i)
∂(x, y)
= r and (ii)
∂(r, θ)
1
∂(r, θ)
= .
∂(x, y)
r
1.1.1
Change of variables
Suppose z = f (x, y) and x = φ(u, v), y = ψ(u, v). Then by chain rule
∂f
∂f ∂x ∂f ∂y
∂f
∂f ∂x ∂f ∂y
=
+
and
=
+
.
∂u
∂x ∂u ∂y ∂u
∂v
∂x ∂v
∂y ∂v
Solving the system of equations by Crammer rule
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Shiv Datt Kumar
Integral Calculus
∂f
∂x
∂f ∂y ∂f ∂y
−
∂u ∂v
∂v ∂u
=
∂f
∂y
∂f ∂x ∂f ∂x
−
∂v ∂u ∂u ∂v
=
1
∂x ∂y ∂x ∂y
−
∂u ∂v
∂v ∂u
∂f
∂f
1
∂x = ∂y =
∂(f, y)
∂(f, x)
∂(x, y)
∂(u, v)
∂(u, v)
∂(u, v)
∂x
∂u
Determinant J = ∂y
∂u
∂x
∂(x, y)
∂v
∂y = ∂(u, v)
∂v
is called the Jacobian of variables of transformation. Therefore
1 ∂(f, y)
∂f
=
∂x
J ∂(u, v)
1 ∂(f, x)
∂f
=−
.
∂y
J ∂(u, v)
and
In case of three variables, let S = f (x, y, z) and x = F (u, v, w), y = G(u, v, w)
& z = H(u, v, w). Then
∂f
∂f ∂x ∂f ∂y
∂f ∂z
=
+
+
.
∂u
∂x ∂u ∂y ∂u ∂z ∂u
∂f ∂x ∂f ∂y ∂f ∂z
∂f
=
+
+
.
∂v
∂x ∂v
∂y ∂v
∂z ∂v
∂f
∂f ∂x
∂f ∂y
∂f ∂z
=
+
+
.
∂w
∂x ∂w
∂y ∂w
∂z ∂w
∂f
1 ∂(f, y, z) ∂f
1 ∂(f, x, z)
∂f
1 ∂(f, x, y)
= [
],
=− [
] and
= [
], where
∂x
J ∂(u, v, w) ∂y
J ∂(u, v, w)
∂z
J ∂(u, v, w)
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂u
∂y
J=
∂u
∂z
∂u
∂x
∂w
∂y
∂w
∂z
∂w
Example 1.1.3. Let z = f (x, y), x = rcos θ, y = rsin θ. Then
∂f 2
∂f
∂f
1 ∂f
) + ( )2 = ( )2 + 2 ( )2 .
∂x
∂y
∂r
r ∂θ
∂(x, y)
Here J =
= r,
∂(r, θ)
∂f
1 ∂(f, y)
=
.................(1) and
∂x
J ∂(r, θ)
(
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Shiv Datt Kumar
Integral Calculus
∂f
1 ∂(f, x)
=−
.................(2).
∂y
J ∂(r, θ)
∂f
∂f
∂(f, y)
= rcos θ
− sin θ
........(3) and
Also
∂(r, θ)
∂r
∂ θ
∂f
∂f
∂(f, x)
= −rsin θ
− cos θ
...........(4)
∂(r, θ)
∂r
∂ θ
Squaring and adding (1) & (2) and using (3) & (4) we get the required.
Cartesian coordinates to cylindrical coordinates
Cylindrical coordinates (r, θ, z) are given by x = r cos θ, y = r sin θ, z = z and
Jacobian of transformation is
∂x
∂r
(x, y, z)
∂y
J = J(
)=
(r, θ, z)
∂r
∂z
∂x
cos θ
= sin θ
0
1.2
−r sin θ
r cos θ
0
∂x
∂θ
∂y
∂θ
∂z
∂θ
∂x
∂z
∂x
∂z
∂z
∂z
z
0 =r
1
Volume of solid of revolution
Let AB be the portion of the curve y = f (x), f (x) > 0, x = a, x = b. Consider
the area bounded by the arc AB of the curve y = f (x), x-axis and the lines x = a
and x = b. Volume of the solid generated by revolving this area about the x-axis
Rb
is V = a πy 2 dx. Similarly volume of the solid generated by revolving this area
Rb
about the y-axis and lines y = c, y = d is V = a πx2 dy.
Example 1.2.1. Find the volume of the solid generated by revolving the finite
region bounded by the curve y = x2 + 1, y = 5 about the line x = 3.
R5
R5
√
√
Solution: V = 1 π(x1 2 − x2 2 )dy = 1 π[(3 + y − 1)2 − (3 − y − 1)2 ]dy =
R5
√
1 12π( y − 1)dy = 64π.
Example 1.2.2. Consider the element δxδy at P (x, y) of plane area A. As this
elementary area revolves about x-axis, we get a ring of volume π[(y + δy)2 −
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Shiv Datt Kumar
Integral Calculus
y 2 ]δx = 2πyδxδy. Total volume=
RR
2
A 2πr sinθ dr dθ.
RR
A
2πy dxdy =
RR
A
2πr sinθ rdr dθ =
Example 1.2.3. Find the volume of the solid generated by revolving the finite
region bounded by the curve y = 3 − x2 , y = −1 about the line y = −1.
Solution: V = 2[ volume of solid generated about the y = −1]
R2
R2
R2
512π
= 0 π(1 + y)2 dx = 0 2π(1 + y)2 dx = 0 2π(1 + 3 − x2 )2 dx =
.
15
Example 1.2.4. Find the volume of the solid generated by revolving the arc of
cycloid x = a(t − sint), y = a(1 − cost), about x−axis.
R 2πa
R 2π
Rπ
Solution: V = 0 πy 2 dx = 0 πa2 (1−cos t)2 a(1−cos t)dt = 16πa3 0 sin6 tdt =
5π 2 a3 .
1.3
Double Integral
Notion of double integral is an extension of the notion of definite integral on the
real line to the case of two dimensional space R2 . Let f (x, y) be a continuous
function in a simply connected, closed bounded region R in two variables x and y.
Divide the region R into subregions (rectangles) by drawing lines x = xk , y = yk ,
k = 1, 2, . . . , m parallel to coordinate axes. Let (xi , yi ) be an arbitrary point
Pn
inside the ith rectangle, whose area is △Ai . Let Sn = i=1 f (xi , yi )△Ai . When
n → ∞, the number of subregions increase indefinitely such that the largest of
Pn
areas △Ai approaches zero. Then limn→∞ Sn = limn→∞ i=1 f (xi , yi )△Ai , if
exists, is called the double integral of the function f (x, y) over the region R and
RR
is denoted by
R f (x, y)dxdy.
Properties of double integrals
If f (x, y) and g(x, y) are integrable functions, then
1.
RR
f (x, y) ± g(x, y)]dxdy =
2.
RR
k f (x, y)dxdy = k
R
3. |
R
RR
R
k f (x, y)dxdy| ≤
RR
R
RR
R
RR
R
f (x, y)dxdy ±
f (x, y)dxdy.
|f (x, y)|dxdy.
ix
RR
R
g(x, y)]dxdy.
Shiv Datt Kumar
Integral Calculus
4. Mean Value Theorem:
RR
R
f (x, y)dxdy = f (u, v)A, where A is the
area of the region R and (u, v) is any arbitrary point in the region R. If
RR
m ≤ f (x, y) ≤ M for all (x, y) in R, then m A ≤
R f (x, y)dxdy ≤ M A.
5. If 0 < f (x, y) ≤ g(x, y) for all (x, y) ∈ R, then
6. If f (x, y) ≥ 0, then
RR
R
Example 1.3.1. Evaluate
R 1 R √1+x2
RR
R
f (x, y)dxdy ≤
RR
R
g(x, y)dxdy.
f (x, y)dxdy ≥ 0.
R 1 R √1+x2
0
0
dydx
.
1 + x2 + y 2
√
R 1 R 1+x2
= 0 dx 0
dydx
dy
2
2
1+x +y
1 + x2 + y 2
√
R1
R
π
y
1
1
2
1
= 0 dx[ √
tan−1
]0 1+x = 0 dx[ √
]
2
2
2
1+x
1+x
1+x 4
√
√
1
π
π R1
π
√
dx = [log(x + x2 + 1]]10 = [log( 2 + 1)].
=
4 0 1 + x2
4
4
RR
xy dx dy, where A is the region bounded by the
Example 1.3.2. Evaluate
A
Solution:
0
0
ordinate x = 2a, curve x2 = 4ay.
x2
Solution: Limit of integration for x are from 0 to 2a and for y from 0 to
.
4a
R 2a R x2 /4a
RR
R 2a
R x2 /4a
xy dx dy = 0 x dx 0
y dy = a4 /3.
A xy dx dy = 0
0
Example 1.3.3. Evaluate
RR
R
2
ex dxdy, where R : 2y ≤ x ≤ 2 and 0 ≤ y ≤ 1.
First integrate with respect to y and then with respect x.
R 2 x2 x2
R 2 R x2 x2
1 R 2 x2
1 4
0 [ 0 e dy]dx = 0 [ye ]0 dx = 2 0 xe dx = 4 (e − 1).
Applications of double integrals
1. If f (x, y) = 1, then
RR
R
dxdy= Area of the region R.
2. If z = f (x, y) is a surface, then
RR
R
f (x, y)dxdy = Volume of the region
beneath the surface z = f (x, y) and above the x − y plane.
3. If f (x, y) = ρ(x, y) is the density function (mass per unit area), then
RR
ρ(x, y)dxdy = M (Total mass of the region R).
(i)
R
(ii) Centre of gravity (x̄, ȳ) is given by
1 RR
1 RR
x̄ =
xρ(x, y)dxdy, ȳ =
R
R yρ(x, y)dxdy.
M
M
RR 2
(iii) Moment of inertia of the mass in R about x-axis, Ix =
R y ρ(x, y)dxdy
RR 2
Moment of inertia of the mass in R about y-axis, Iy =
R x ρ(x, y)dxdy
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Shiv Datt Kumar
1.3.1
Integral Calculus
Change of variables in double and triple integral
Let R be the domain of integration
RR
R
f (x, y) dxdy in x − y plane Let x =
φ(u, v), y = ψ(u, v) and R∗ be the domain in the u − v plane. Then
RR
RR
R∗ F (u, v) |J| dudv, where
R f (x, y) dxdy=
∂x
∂u
Determinant J = ∂y
∂u
∂x
∂(x, y)
∂v
∂y = ∂(u, v) .
∂v
If x = φ(u, v, w), y = ψ(u, v, w),z = η(u, v, w), then
RRR
RR
R∗ F (u, v, w) |J| dudvdw, where
R f (x, y, z) dxdydz =
∂x
∂u
∂y
J=
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
Example 1.3.4. Let R be rhombus with successive vertices at (1, 0), (2, 1), (1, 2)
RR
2
2
and (0, 1). Then find the value of the integral
R (x − y) cos (x + y)dxdy.
Solution: Equations of sides AB, BC, CD, and DA are given by x − y = 1,
x + y = 3, x − y = −1 and x + y = 1 respectively. Put x − y = u, y + x = v.
Then −1 ≤ u ≤ 1, 1 ≤ v ≤ 3 and x = (v − u)/2, y = (u + v)/2. Jacobian of
transformation is given by
∂x
∂(x, y)
J=
= ∂u
∂y
∂(u, v)
∂u
∂x
∂v
∂y = −1/2
∂v
R3R1
RR
(x − y)2 cos2 (x + y) dxdy = 1 −1 u2 cos2 v|J| dudv
Then I =
R
1 R3
1 R3
1 R3R1 2 2
cos2 vdv =
(1 + cos 2v)dv
u cos v dudv =
=
2 1 −1
3 1
6 1
1 sin 6 − sin 2
= +
.
3
12
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Shiv Datt Kumar
1.4
Integral Calculus
Triple Integral
Triple integral is an extension of the notion of double integral to three dimensional space R3 . Let f (x, y, z) be a continuous function in a simply connected,
closed bounded volume V . Divide the region V into small volume elements by
drawing planes x = xk , y = yk , z = zk , k = 1, 2, . . . , n parallel to three coordinate planes. Let (xi , yi , zi ) be an arbitrary point inside the ith volume element
P
δVi = δxi δyi δzi . Let Sn = ni=1 f (xi , yi , zi )δVi . When n → ∞, the number of
subregions increase indefinitely such that the largest of volumes δVi approaches
Pn
zero. Then limn→∞ Sn = limn→∞ i=1 f (xi , yi , zi )δVi , if exists, is called the
triple integral of the function f (x, y, z) over the region V and is denoted by
RRR
V f (x, y, z)dxdydz.
Let f (x, y, z) be a continuous function over a regular solid V defined by
a < x < b, h1 (x) < y < h2 (x), and g1 (x, y) < z < g2 (x, y). Then the triple
integral is equal to the triple iterated integral given by
RRR
V
f (x, y, z)dxdydz =
R b R h2 (x) R g2 (x,y)
a
h1 (x)
g1 (x,y)
f (x, y, z)dxdydz.
Applications of triple integrals
1. If f (x, y, z) = 1, then
RRR
R
dxdydz= Volume of the region R.
2. If f (x, y, z) = ρ(x, y, z) is the density of mass, then
RRR
(i)
R ρ(x, y)dxdydz = M (Total mass of the solid of region R).
(ii) Centre of gravity (x̄, ȳ, z̄) is given by
1 RRR
1 RR
x̄ =
R x ρ(x, y, z)dxdydz, ȳ = M
R y ρ(x, y, z)dxdydz,
M
1 RR
z̄ =
R z ρ(x, y, z)dxdydz.
M
(iii) Moment of inertia of the mass in R about x-axis,
RRR
(y 2 + z 2 )ρ(x, y, z)dxdydz
Ix =
R
Moment of inertia of the solid of mass in R about y-axis,
RRRR
(x2 + z 2 )ρ(x, y, z)dxdydz and
Iy =
R
moment of inertia of the solid of mass in R about z-axis,
RRRR
(x2 + y 2 )ρ(x, y, z)dxdydz.
Iz =
R
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Shiv Datt Kumar
Integral Calculus
Example 1.4.1. Find the volume of the solid bounded by the plane x = 0, y = 0,
x + y + z = a and z = 0.
R a R a−x R a−x−y
R a R a−x
1 Ra
(a −
Volume V = 0 0
dx dy dz = 0 0 (a − x − y)dy dx =
0
2 0
x)2 dx = a3 /6.
Example 1.4.2. Find the volume generated by the revolution of cardioid
r = a(1 − cos θ) about its axis.
R π R a(1−cos θ)
Volume V = 0 0
2πr2 drsin θdθ
3 R
2πa π
πa3
3
=
(1 − cos θ) sin θdθ =
.
0
3
6
Example 1.4.3. Calculate the volume of the ellipsoid
x2
y2
z2
+
+
= 1.
a2
b2
c2
Using the transformation of spherical coordinates x = arsinθ cosφ, y =
brsinθ sinφ, z = crcosθ. Jacobian of transformation
∂x
∂r
∂y
J=
∂r
∂z
∂r
∂x
∂θ
∂y
∂θ
∂z
∂θ
∂x
∂φ
∂y
= abc r2 sin θ
∂φ
∂z
∂φ
dxdydz = Jdr dθ dφ = abc r2 sin θdr dθ dφ
V = 8× volume of sphere in positive octant
R 1 R π/2 R π/2
=8 0 0
abc r2 sin θdrdθ dφ
0
R1
R π/2
R π/2
4
= 8 0 abc r2 dr 0 sin θdθ 0
dφ = πabc.
3
Example 1.4.4. Calculate the volume of the solid surrounded by the surface
(x/a)2/3 + (y/b)2/3 + (z/c)2/3 = 1.
Using the transformation of coordinates x = aX 3 , y = bY 3 , z = cZ 3 . Ja∂(x, y, z)
cobian of transformation J =
= 27abcX 2Y 2 Z 2 . Required volume =
∂(X, Y, Z)
RRR
RRR 2 2 2
dx dy dz = 27abc
X Y Z dX dY dZ
Now use the transformation X = rsin θcos φ, Y = rsin θsin φ, Z = rcos θ.
∂(X, Y, Z)
′
Then Jacobian of transformation J =
= r2 sin θ drdθdφ . Thus
∂(r, θ, φ)
dXdY dZ = r2 sin θdrdθdφ and dxdydz = 27abcX 2Y 2 Z 2 r2 sin θ drdθ dφ .
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Shiv Datt Kumar
Integral Calculus
V = 8× volume in positive octant=
R 1 R π/2 R π/2
8 0 0
27abc (rsin θ cos φ)2 (rsin θ sin φ)2 (rcos θ)2 r2 sin θdrdθ dφ.
0
R1
R π/2
R π/2
4πabc
= 216abc 0 r8 dr 0 sin5 θcos2 θdθ 0 sin2 φ cos2 φ dφ =
.
35
Example 1.4.5. Find total mass of the solid bounded by the surfaces x2 + y 2 =
16, z = 2 and z = 4 if its density function is ρ(x, y, z) = x2 + y 2 .
R z=4
R4
R √16−x2
Convert the integral x=−4 y=−√16−x2 z=2 ρ(x, y, z) dx dy dz in cylindrical co-
ordinates x = rcos θ, , y = rsin θ, z = z, then Jacobian J = r and
R 4 R 2π R 4
R4
R 2π
R4
r4 4
2
3
} ×2π×2 = 256π.
r
r
dr
dθ
dz
=
r
dr
dθ
dz
=
{
r=0 θ=0 z=2
r=0
θ=0
z=2
4 0
Example 1.4.6. Show that in the first octant, paraboloid z = 36 − 4x2 − 9y 2 has
the volume 27π cubic units.
Solution: Projection of the given paraboloid in the x y plane is the first quadrant
of the ellipse 4x2 + 9y 2 = 36. Then region R is given by 0 ≤ z ≤ 36 − (4x2 + 9y 2 ),
1√
0≤y≤
36 − 4x2 , 0 ≤ x ≤ 3. Therefore volume
3 √
R 3 R 2 (9−x2 )
V = 0 [ 03
(36 − 4x2 − 9y 2 )dy]dx
√
2
R3
16 R 3
(9−x2 )
(9 − x2 )3/2 dx.
= 0 [4(9 − x2 )y − 3y 3 ]03
dx =
9 0
R π/2
3 1 π
Now put x = 3sin θ. Then V = 144 0 cos4 θdθ = 144( . . ) = 27π cubic
4 2 2
unit.
Example 1.4.7. Find the volume of the solid which is contained in between the
cone z 2 = 2(x2 + y 2 ) and the hyperboloid z 2 = x2 + y 2 + a2 .
Solution: Put x = rcos θ, y = rsin θ. Then dxdy = rdrdθ and intersection of
z 2 = 2(x2 +y 2 ) and z 2 = x2 +y 2 +a2 is 2(x2 +y 2 ) = x2 +y 2 +a2 i.e. x2 +y 2 = a2 .
Therefore
p
R 2π R a R √x2 +y2 +a2
R 2π R a p
V = 0 0 √ 2 2 dz dy dx = 0 0 (x2 + y 2 + a2 )− 2(x2 + y 2 ) dy dx =
2(x +y )
√
√
R 2π R a p
4πa3 ( 2 − 1)
2 + a2 ) −
(r
2r)r
dr
dθ
=
.
(
0
0
3
Example 1.4.8. Find mass of a solid lying between spheres of radius 3 and 4
in the region y ≥ 0, z ≥ 0. Density at any point (x, y, z) is given by ρ(x, y, z) =
p
x2 + y 2 + z 2 .
Solution: Convert the integral into spherical coordinates x = r sin θ cos φ,
xiv
Shiv Datt Kumar
Integral Calculus
y = r sin θ sin φ, z = r cos θ, Region occupied by the solid is described by
π
3 ≤ r ≤ 4, 0 ≤ θ ≤ π, 0 ≤ φ ≤ . Also
2
ρ(r, θ, φ) = ρ(r sin θ cos φ, r sin θ sin φ, r cos θ)
p
= (r sin θ cos φ)2 + (r sin θ sin φ)2 + (r cos θ)2 = r.
Jacobian of transformation J = r2 sin φ.
R π R π/2 R 4 2
Rπ
R π/2
R4
Hence mass= 0 0
r r sin φ dr dφ dθ = 0 dθ 0 sin φ dφ 3 r3 dr =
3
175π
4
RRR
Exercise 1.4.1.
1. Evaluate
sin(x + y + z) dx dy dz over portion cut off
by the plane x + y + z = π.
x−y
dx dy,
x+y
where R is the triangular region bounded the vertices (1, 0), (4, −3), (4, 1).
1
75
Ans: (49 log 7 − log 5 − 27 log 3 + 6).
4
2
RR
2
3. Evaluate
R (x + y) dx dy, where R is the region bounded by the ellipse
2
2
y
x
+ 2 = 1, using transformation of variables.
2
a
b
πab 2
(a + b2 ).
Ans:
4
2. By a suitable change of variables calculate the integral
4. Show that the area of the surface of paraboloid
cylinder
x2
y2
2
+
= k is π{(1 + k)3/2 − 1}ab.
a2
b2
3
RR
R
log
y2
x2
+
= 2z inside the
a
b
5. Find the area of the part of the cylinder x2 + y 2 = a2 which is cut by the
cylinder x2 + z 2 = a2 .
xy(x2 − y 2 )
, (x, y) 6= (0, 0), f (0, 0) = 0. Then show that
x2 + y 2
fxy (0, 0) 6= fyx (0, 0).
6. Let f (x, y) =
xv
Shiv Datt Kumar
Beta and Gamma Functions
xvi
Chapter 2
Improper Integrals, Beta
and Gamma Functions
2.1
Improper Integral
For the existence of Riemann integral (definite integral)
Rb
a
f (x)dx, we require
that the limit of integration a and b are finite and function f (x) is bounded. In
case
(i) limit of integration a or b or both become infinite (improper integral of first
kind),
(ii) integrand f (x) has singular points (discontinuity) i.e. f (x) becomes infinite
at some points in the interval a ≤ x ≤ b (improper integral of second kind),
Rb
then the integral a f (x)dx is called improper integral. Note that improper
integral are evaluated by limiting process.
Example 2.1.1.
R ∞ −1
R∞
R1
1
dx
−1 x2 dx, −∞ 1 + x2 dx, 0 x(1 − x) are improper integrals.
Definition 2.1.2. If a is the only point of infinite discontinuity of f (x), then
Z
a
b
f (x)dx = lim
ǫ→0+
Z
b
f (x)dx (if f inite limit exists)
a+ǫ
xvii
Shiv Datt Kumar
Beta and Gamma Functions
If b is the only point of infinite discontinuity, then
Z
b
f (x)dx = lim
ǫ→0+
a
Z
b−ǫ
f (x)dx,
a
0<ǫ<b−a
If end points a and b are the only points of infinite discontinuity,
Z
b
f (x)dx =
a
lim
ǫ→0+ & µ→0+
Z
b−µ
f (x)dx,
a−ǫ
Improper integral is convergent if limit exists and is finite otherwise it is divergent.
If an interior point c,
a < c < b, is the only point of infinite discontinuity,
then we write
Rb
Rc
Rb
f (x)dx = a f (x)dx + c f (x)dx.
a
Improper integral is convergent if both the limits exist and are finite otherwise
it is divergent.
Example 2.1.3. Examine the convergence of
Rb
a
dx
(x − a)n
Solution: It is a proper integral if n ≤ 0 and improper for other values of n.
For n 6= 1,
Rb
Rb
dx
dx
= limp→0+ a+p
,
a (x − a)n
(x − a)n
0<p<b−a
1
,
1
1
1
(1 − n)(b − a)n−1
= lim
−
]
=
[
p→0+ (1 − n) (b − a)n−1

pn−1

 ∞




For n = 1
Rb
Rb
dx
dx
= limp→0+ a+p
a (x − a)n
(x − a)
= limp→0+ log(b − a) − log p = ∞
Thus integral converges only if n < 1.
Comparison Test I
xviii
if n < 1
if n > 1,
Shiv Datt Kumar
Beta and Gamma Functions
If f (x) and g(x) are two functions such that f (x) ≤ g(x), for all x ∈ [a, b],
then
1.
Rb
f (x) dx converges if
2.
Rb
g(x) dx diverges if
a
a
Rb
a
Rb
a
g(x) dx converges.
f (x) dx diverges.
Comparison Test II
f (x)
If f (x) and g(x) are two functions such that limx→∞
= l (nonzero & finite),
g(x)
R∞
R∞
then a f (x) dx and a g(x) dx converge or diverge together.
Example 2.1.4. Examine the convergence of (i)
(ii)
R∞
R π/2 sin x dx
dx
(iii) 1 3 −x
0
n
x
x (e + 1)
R1
0
dx
p
(1 − x3 )
1
1
p
Solution: (i) Let f (x) = p
= p
.
(1 − x3 )
(1 − x) (1 + x + x2 )
1
is a bounded function and let M be its upper bound.
Note that p
(1 + x + x2 )
R1
M
1
Then f (x) ≤
and 0
dx is convergent. Therefore by com1/2
(1 − x)
(1 − x)1/2
R1
dx
parison test 0 p
is convergent.
(1 − x3 )
(ii) For n ≤ 1, it is a proper integral. For n > 1, it is an integral and 0 is the point
sin x
sin x 1
sin x
of infinite discontinuity. Now
=
. Function
is bounded
n
n−1
x
x x
x
R
1
sin x
sin x
π/2 dx
≤ n−1 and 0
converges only if n − 1 < 1
≤ 1. Thus
and
x
xn
x
xn−1
or n < 2 and diverges for n ≥ 2.
1
1
(iii) Let f (x) = 3 −x
and g(x) = 3 . Also
x (e + 1)
x
x3
1
1
f (x)
= lim 3 −x
= lim −x
= 1.
x→∞ x (e
x→∞ e
x→∞ g(x)
+ 1) 1
+1
lim
R∞ 1
R∞
dx
dx converges. Therefore 1 3 −x
converges.
1 x3
x (e + 1)
R π/2
R π/2 cosn x
dx,
Example 2.1.5. Examine the convergence of (i) −π/2 tan x dx (ii) 0
xm
if m < 1.
Also
Solution:
(i)
Z
π/2
−π/2
tan x dx = lim
ǫ→0
Z
c
tan x dx + lim
−π/2+ǫ
xix
η→0
Z
c
π/2−η
tan x dx
Shiv Datt Kumar
Beta and Gamma Functions
= lim ln[cos (−π/2 + ǫ)] − ln[cos c] − lim ln[cos (π/2 − η)] − ln[cos c].
ǫ→0
η→0
Limits do not exist. Hence improper integral diverges.
cosn x
1
(ii) Note that
< m for 0 < x < π/2. Also x = 0 is the point of infinite
xm
x
R π/2 1
discontinuity and 0
dx is convergent if m < 1 by comparison test.
xm
Absolute Convergence of Improper Integrals
If the function f (x) changes sign within the interval of integration, we consider
Rb
absolute convergence. The improper integral a f (x) dx is called absolutely conRb
vergent if a |f (x)| dx converges. Since f (x) ≤ |f (x)|, ∀ x, absolutely convergent
improper integral is convergent.
R ∞ sin x
Example 2.1.6. Examine the convergence of improper integral −∞
dx.
1 + x2
R ∞ sin x
R ∞ sin x
| dx ≤ −∞ |
| dx
Solution: Note that |I| = | −∞
1 + x2
1 + x2
R b sin x
R c sin x
| dx + limb→∞ c |
| dx = I1 + I2 .
= lima→−∞ a |
1 + x2
1 + x2
Rc
R c sin x
1
π
| dx ≤ lima→−∞ a |
| dx = tan−1 c + .
I1 = lima→−∞ a |
2
2
1+x
1+x
2
Rb
R b sin x
1
π
−1
| dx ≤ limb→∞ c |
| dx = − tan c.
I2 = limb→∞ c |
1 + x2
1 + x2
2
Thus |I| ≤ |I1 | + |I2 | ≤ π and hence convergent.
2.2
Beta Function
The improper integral defined by
R1
β(m, n) = 0 xm−1 (1 − x)n−1 dx for m > 0, n > 0 is called as beta function.
Note that
R 1 m−1
(1 − x)n−1 dx converges if m > 0, n > 0.
0 x
Properties:
1. β(m, n) = β(n, m)
Put x = 1 − y. Then
β(m, n) = −
2. β(m, n) =
R0
1
R π/2
0
(1 − y)m−1 y n−1 dy =
R1
0
y n−1 (1 − y)m−1 dy = β(n, m).
sin2m−1 θcos2n−1 θ dθ, n > 0.
Put x = sin2 θ, dx = 2sin θ cos θ dθ in β(m, n) =
xx
R1
0
xm−1 (1 − x)n−1 dx.
Shiv Datt Kumar
Beta and Gamma Functions
3. β(m, n) = β(m + 1, n) + β(m, n + 1)
2.2.1
Gamma Functions
R∞
The improper integral of the form Γ(n) =
0
function.
e−x xn−1 dx, n > 0 is called gamma
Properties:
1. Γ(1) =
R∞
0
e−x dx = 1.
2. Reduction formula Γ(n + 1) = nΓ(n)
Proof: Γ(n + 1) =
R∞
0
e−x xn dx = [−xn e−x ]∞
0 +n
R∞
0
e−x xn−1 dx
= nΓ(n) = n!.
√
π.
3. Γ(1/2) =
R∞
Proof: Γ(1/2) =
2
(Γ(1/2)) = 4
=4
R π/2 R ∞
0
0
0
e−x x1/2 dx = 2
R∞R∞
0
0
e−(x
2
e−r rdrdθ =
2
+y 2 )
R∞
0
2
e−y dy (by putting x = y 2 ).
dxdy
4π R ∞ −r2
e rdr = π
2 0
4. Relation between β and Γ functions
Γ(m)Γ(n)
Γ(m + n)
R∞
R∞
2
Proof: Γ(m) = 0 e−t tm−1 dt = 2 0 e−x x2m−1 dx
β(m, n) =
Γ(n) = 2
R∞
0
2
e−y y 2n−1 dy
Γ(m)Γ(n) = 4
=4
R∞R∞
0
0
R π/2 R ∞
=2
R∞
0
0
e−(x
=4
0
R∞
0
2
2
e−x x2m−1 dx
R∞
+y 2 ) 2m−1 2n−1
x
y
0
2
e−y y 2n−1 dy
dxdy
2
e−r r2(m+n)−1 cos2m−1 θ sin2n−1 θ drdθ
2
e−r r2(m+n)−1 dr × 2
R π/2
0
cos2m−1 θ sin2n−1 θdθ
= Γ(m + n)β(m, n).
5.
R π/2
0
cosm θ sinn θdθ
m+1
n+1
Γ(
)Γ(
)
1 m+1 n+1
2
2
,
)=
= β(
m+n+2
2
2
2
2Γ(
)
2
xxi
Shiv Datt Kumar
Beta and Gamma Functions
R ∞ xm−1
π
Example 2.2.1. If 0
dx =
, 0 < m < 1, then Γ(m)Γ(1 − m) =
1−x
sin mπ
π
.
sin mπ
x
y
1
dy. Thus
Put
= y. Then x =
and dx =
1+x
1−y
(1 − y)2
m−1
R1
R∞ x
dx = 0 y m−1 (1 − y)m−1 dy = β(m, 1 − m)
0 1−x
R ∞ xm−1
Γ(m)Γ(1 − m)
π
=
= Γ(m)Γ(1 − m). Hence 0
dx =
.
Γ(m + 1 − m)
1−x
sin mπ
Example 2.2.2. (Legendre Duplication formula )
√
1
π Γ(2m) = 22m−1 Γ(m)Γ(m + ).
2
R π/2
Γ(m)Γ(n)
Since
= β(m, n) = 2 0 sin2m−1 θ cos2n−1 θ dθ...............(1)
Γ(m + n)
Putting n = m, we have
2
R π/2
Γ(m)
= β(m, m) = 2 0 sin2m−1 θ cos2m−1 θ dθ.
Γ(2m)
1 R π/2
= 2m−2 0 sin2m−1 2θ dθ. Put 2θ = φ. Then
2
1 R π/2
= 2m−1 0 sin2m−1 φ dφ........................(2).
2
In (1) take n = 1/2, we get
R π/2
Γ(m)Γ(1/2)
= 2 0 sin2m−1 θdθ...............(3).
Γ(m + 1/2)
From (2) and (3), we get,
2
√
1 Γ(m)Γ(1/2)
Γ(m)
= 2m−1
. Since Γ(1/2) = π, we have
Γ(2m)
2
Γ(m + 1/2)
√
1
2m−1
π Γ(2m) = 2
Γ(m)Γ(m + ).
2
1
π
1
.
Example 2.2.3. β(m + , m + ) =
2
2
m24m−1 β(m, m)
R 1 m−1
Solution: By definition β(m, n) = 0 x
(1 − x)n−1 dx, m > 0, n > 0. By
R π/2
substituting x = sin2 θ, we have β(m, n) = 2 0 sin2m−1 θ cos2n−1 θ dθ.
1
First show that β(m, m) = 2m−1 β(m, 1/2) ......................(1)
2
√
π
...............(2)
Then Γ(m + 1/2)Γ(m) = 2m−1 Γ(2m)
2
R
Γ(m)Γ(n)
π/2
= β(m, n) = 2 0 sin2m−1 θ cos2n−1 θ dθ, we have
Since
Γ(m + n)
Γ(m + 1/2)Γ(m + 1/2)
. Using (2) we have
β(m + 1/2, m + 1/2) =
Γ(2m + 1)
1
1
π
β(m + , m + ) =
.
2
2
m24m−1 β(m, m)
R b(α)
Example 2.2.4. (i) Let φ(α) = a(α) f (x, α)dx. Then by Leibniz formula
dφ R b(α) ∂f (x, α)
db
da
dφ
π
= a(α)
dx + f (b, α)
− f (a, α)
show that
=
.
dα
∂α
∂α
∂α
da
2(a + 1)
xxii
Shiv Datt Kumar
Beta and Gamma Functions
R ∞ tan−1 (ax)
dx.
0
x(1 + x2 )
R ∞ tan−1 (ax)
Solution: Let φ(a) = 0
dx. Then by Leibniz formula
x(1 + x2 )
−1
R∞
dx
dφ R ∞ ∂ tan (ax)
= 0
[
]dx = 0
2
2
da
∂ a x(1 + x )
(1 + x )(1 + a2 x2 )
2
R
a
1
1
1
∞
−1
[
]dx = 2
−
[{a tan−1 (ax)}∞
x}∞
= 2
0 − {tan
0 ]
a − 1 0 a2 x2 + 1 1 + x2
a −1
π a−1
= [ 2
], a > 0, a 6= 1
2 a −1
π
. Integrating with respect to a, we have
=
2(a + 1)
π
φ(a) = ln(a + 1) + c. Since φ(0) = 0, c = 0.
2
R ∞ tan−1 (ax)
π
dx = ln(a + 1).
Therefore φ(a) = 0
2
x(1 + x )
2
(ii) φ(a) =
2.3
Dirichlet Integral
Theorem 2.3.1.
xl−1 y m−1 dxdy =
RR
D
Γ(l)Γ(m) l+m
h
, where D is domain
Γ(l + m + 1)
x ≥ 0, y ≥ 0, x + y ≤ h.
Proof. Put x = Xh, y = Y h, dxdy = h2 dXdY . Then
RR
R R l−1 m−1
l−1
(Y h)m−1 dXdY
y
dxdy =
D (Xh)
Dx
R 1 R 1−X l−1 m−1
= hl+m 0 0
X Y
dXdY
R
R
1
1−X
= hl+m 0 X l−1 dX 0
Y m−1 dY
m
R1
Y 1−X
= hl+m 0 X l−1 dX[
]
m 0
l+m R
h
1
=
X l−1 (1 − X)m dX
m 0
hl+m
=
β(l, m + 1)
m
hl+m Γ(l)Γ(m + 1)
=
m Γ(l + m + 1)
l+m Γ(l)Γ(m)
=h
Γ(l + m + 1)
Theorem 2.3.2.
RRR
V
xl−1 y m−1 z n−1 dxdydz =
Γ(l)Γ(m)Γ(n)
, where V is
Γ(l + m + n + 1)
region x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1.
Proof. Put x + z ≤ 1 − x = h. then z ≤ h − y.
R1
R 1−x
R 1−x−y n−1
R R R l−1 m−1 n−1
y
z
dxdydz = 0 xl−1 dx 0 y m−1 dy 0
z
dz
V x
R 1 l−1 R h R h−y m−1
= 0 x dx[ 0 0
y
dyz n−1 dz]
xxiii
Shiv Datt Kumar
Beta and Gamma Functions
=
R1
0
xl−1 dx[hm+n
Γ(m)Γ(n)
]
Γ(m + n + 1)
Γ(m)Γ(n) R 1 l−1
x (1 − x)m+n dx
Γ(m + n + 1) 0
Γ(m)Γ(n)
=
β(l, m + n + 1)
Γ(m + n + 1)
Γ(m)Γ(n) Γ(l)Γ(m + n + 1)
=
Γ(m + n + 1) Γ(l + m + n + 1)
Γ(l)Γ(m)Γ(n)
=
Γ(l + m + n + 1)
=
Corollary 2.3.3.
RRR
V
xl−1 y m−1 z n−1 dxdydz =
Γ(l)Γ(m)Γ(n) l+m+n
h
, where
Γ(l + m + n + 1)
V is region x ≥ 0, y ≥ 0, z ≥ 0 x + y + z ≤ h.
Exercise 2.3.4.
1. Discuss the convergence of (i)
R2
1
√
R π/2 sin x
x
√ dx.
dx (ii) 0
ln x
x x
2. Examine the convergence of improper integrals (i)
R ∞ dx
R∞
R∞
dx
2
.
(iv) 1
(ii) 1 e−x dx (iii) 1 2 −x
x (e + 1)
xp
R ∞ dx
2 ln x
3. Show that for m, n > 0,
R∞
t
xm−1
dx. Hint: Put x =
β(m, n) = 0
(1 + x)m+n
1−t
4. Show that
R ∞ xc
Γ(c + 1)
dx =
.
0 cx
(log c)c+1
√
22n−1
5. Show that Γ(2n) = √ Γ(n + 12 )Γ(n) and hence Γ(1/4)Γ(3/4) = π 2.
π
√
R π/2 √
R π/2 √
1 1
3
tan θ dθ = Γ( )Γ( ) = π 2. (Hint: 0
tan θ dx =
0
2 4
4
√
R π/2
π 2
π
1 Γ(1/4)Γ(3/4)
=
= √ .
sin1/2 θ cos−1/2 θ dθ =
0
2
Γ(1)
2
2
R ∞ √ −x2
7. Evaluate the improper integral 0
xe
dx.
6. Show that
xxiv
Bibliography
[1] R K Jain and S R K Iyengar, Advance Engineering Mathematics, Narosa
Publishing Corporation.
[2] Ervin Kreszig, Advance Engineering Mathematics.
[3] S C Malik, Savita Arora, Mathematical Analysis, New Age International
Publishers, Fifth edition, 2017.
xxv
Index
Change of variables, vi, xi
Gamma function, xxi
Jacobian, v, vi
Legendre Duplication formula, xxii
Volume of solid of revolution, viii
xxvi