17 Evaluation Of Improper Integrals

Chapter 7: Evaluation of Improper Integrals • Advice 1: • Page No. 257:

Q. Nos.: 1 - 5

(1) Let f(x) is continuous for all x ≥ 0, then ∞

∫

R

∫ R →∞

f ( x)dx = lim

0

f ( x) dx

0

provided the limit on RHS exists.

( 2) Let f(x) is continuous for all x. ∞

then

∫

f ( x) dx

-∞ 0

= lim

∫

R1→∞ − R1

R2

∫ R2 →∞

f ( x) dx + lim

f ( x)dx,

0

provided both the limits on RHS exist.

Cauchy principal value (P.V.) of the integral (2) is the number ∞

P.V. ∫ f(x) dx = lim -∞

R →∞

R

∫

-R

f ( x) dx,

provided the limit on RHS. exist.

Remark : (1) Existence of improper integral ∞

∫-∞

f ( x) dx implies the existence of ∞

P.V. ∫ f ( x) dx ∞

But converse is not true.

Ex. Let f(x) = x.Then ∞

P.V. ∫ f ( x ) dx = lim -∞

R →∞

R

x  = lim   = 0 R →∞ 2   −R 2

∫

R

xdx

−R

∞

But

∫

f ( x) dx

−∞

= lim

0

xdx + lim ∫

R1→∞ − R1

= − lim

R1→∞

2 R1

2

R2

xdx ∫

R2 →∞ 0 2 R2

+ lim

R2 →∞

2

∴ Limit on RHS fails to exist ⇒ The improper integral ∞

f ( x ) dx fails to exist . ∫

−∞

If the function f(x) ( − ∞ < x < ∞) is an even function i.e. f(-x) =f(x) for all x, then the symmetry of the graph of y = f(x) with respect to y axis leads to R

∫

−R

R

f ( x)dx = 2 ∫ f ( x)dx 0

When f(x) is an even function and the Cauchy pricncipal value exists, then ∞

∞

−∞

−∞

P.V . ∫ f ( x)dx =

∫

∞

f ( x)dx = 2 ∫ f ( x)dx 0

To evaluate improper integral of Even Rational Functions f(x)=p(x)/q(x) • p(x) and q(x) are polynomials with real coefficients and no factors in common • q(z) has no real zeros but has at least one zero above the real axis.

Method • Identify all distinct zeros of the polynomial q(z) that lie above the real axis • They will be finite in number • May be labeled as z1, z2,…..zn where n is less than or equal to the degree of q(z) • Now, integrate the quotient f(z)=p(z)/q(z) around the positively oriented boundary of the semicircular region.

The simple consists of

closed

contour

• The segment of the real axis from z = -R to z =R and • The top half of the circle z = R described counterclockwise and denoted by CR.

Remark: •The positive number R is large enough that the points z1, z2,… zn all lie inside closed path.

. CR

zi

-R

z1

O

R

From Cauchy Residue theorem R

∫

−R

f ( x)dx +

∫

CR

n

f ( z )dz = 2πi ∑ Res f ( z ) z = z k k =1

If lim

R →∞

∫

f ( z )dz = 0,

CR

then it follows ∞

n

P.V . ∫ f ( x)dx = 2πi ∑ Res f ( z ) −∞

z = z k k =1

If f(x) is even, then ∞

n

−∞

k =1

f ( z) ∫ f ( x)dx = 2πi ∑ zRes =z k

and ∞

∫ 0

n

f ( x)dx = πi ∑ Res f ( z ) z = z k k =1

∞

Q.4 Evaluate I = ∫ 0

2

(

x dx 2 2 x +1 x + 4

)(

)

cR

-R

R

Let f(z) =

z

2

( z + 1)( z 2

2

+4

)

& C = [ − R, R ]U C R

then R

∫

−R

2

x dx

( x + 1)( x 2

2

+4

)

+

∫ f ( z ) dz = 2πi ∑

CR

Re s z = zk

f ( z)

clearly z = ±i, ± 2i are poles of order of 1 of f(z) but z = -i, - zi lie outside the region C.

∴Re s f ( z ) = z =i

z

2

( z + i ) ( z + 4) 2

2

i i = = 2i × 3 6

z =i

2

z Re s f ( z ) = 2 z = 2i z + 1 ( z + 2i ) −4 −i = = − 3 × 4i 3

(

)

z = 2i

∴ (1) ⇒ R

∫(

−R

2

x dx i i + f ( z ) dz = 2 π i −   ∫ 2 2 x +1 x + 4 C  6 3

)(

)

R

=π /3

f ( z) =

z 2

2 2

z +1 z + 4

≤

z

2

( z − 1)( z − 4) 2

2

f ( z ) = • …

z

.

2

z +1 z + 4 2

2

≤

(z

z

2

2

− 1)( z − 4 ) 2

2

R = 2 2 (R − 1)( R − 4) on z = Re

iθ

Hence R .π R ∫ f ( z ) dz ≤ 2 2 C ( R − 1)( R − 4) (M L inequality) as L = πR 2

R

π = → 0 as R → ∞ 1  4   R1 − 2 1 − 2   R  R 

which yields ∞

∫(

−∞

x dx π = 2 2 3 x +1 x + 4 2

∞

⇒ 2∫ 0

∞

⇒∫ 0

(

)(

)

2

x dx

( x + 1)( x 2

2

2

+4

)

=π 3

π = 2 2 x +1 x + 4 6 x dx

)(

)

Sec 73 & 74 Advice 2: Page No. 265, Q. Nos.: 1 and 6

Evaluation of improper integral of ∞ ∞ f ( x ) cos ax dx form ∫− ∞ and ∫− ∞ f ( x) sin ax dx R

∫ f ( x) e

iax

−R

R

R

−R

−R

dx = ∫ f ( x) cos ax dx + i ∫ f ( x) sin ax dx

( together with the fact that the modulus e

iaz

=e

ia ( x +iy )

=e

− ay

.e

iax

⇒e

iaz

=e

− ay

is bounded in the upper half plane y ≥ 0

cos x dx Q.1 I = ∫ 2 , a > b > 0 2 2 2 − ∞ ( x + a )( x + b ) ∞

Let f ( z ) =

(z

1

+ a )( z + b & C = [ − R, R ] ∪ C R 2

2

2

2

)

cR -R

R

f ( z ) has singularity at z = ± ai, ± bi out of which z = ai, bi are inside C & they are simple poles

Re s f ( z ) e = iz

z = ai

−a

e

iz

( z + ai ) ( z + b 2

−a

2

e ie = = − 2 2 2 2 2ai ( b − a ) 2a ( b − a )

)

z = ai

iz

e Re s f ( z )e = 2 2 z = bi ( z + a )( z + bi ) −b −b e ie = = 2 2 2 2 2bi ( a − b ) 2b( b − a ) iz

z = bi

ix

e dx iz ∴∫ 2 + f ( z ) e ∫ 2 2 2 − R ( x + a )( x + b ) C R

= 2πi ∑ Re s ( f ( z )e

2πi.i = 2 2 2 b −a

(

−b

)

iz

)

R

−a  e −b e   −  b  a   −a

π e e    = 2 − 2   a  a −b  b

Taking real parts, we get R

∫(

−R

cos x dx iz + Re f ( z ) e ∫ 2 2 2 2 x +a x +b CR

)(

)

π e e   = 2 2  −  a  a −b  b −b

−a

(1)

∴e =e & on C R iz

f ( z) =

−y

(z

≤1 ∀ y ≥ 0

1

+ a )( z + b 1 ≤ 2 2 2 2 ( R − a )( R − b ) 2

2

2

2

)

∴ Re ∫ e f ( z ) dz ≤

( ) e f z dz ∫

iz

CR

iz

CR

πR ≤ 2 2 2 2 R −a R −b =

π 3

(

)(

a  b  R 1 − 2 1 − 2   R  R  2

2

)

→ 0 as R → ∞

∴(1) yields ∞

∫(

−∞

cos x dx 2 2 2 2 x + a x +b

)(

)

e e    = 2 − 2   a  a −b  b

π

−b

−a

Jordan' s Lemma : Suppose that (i) a function f(z) is analytic at all points z in the upper half plane y ≥ 0 that are exterior to the circle z = R 0 ;

iθ

(ii ) C R : z = R e , 0 ≤ θ ≤ π , R > R0 ; (iii ) for all points z on C R , there is a positive constant M R such that f ( z ) ≤ M R , where lim M R = 0.

R→∞

Then, for every positive constant a, lim

R →∞

∫

CR

f ( z) e

iaz

dz = 0

3

x sin ax Q .6 I = ∫ dx , a > 0 4 −∞ x + 4 ∞

3 iaz

z e Let f ( z ) = 4 z +4

z + 4 = 0 ⇒ z = −4 4

4

⇒ z = ( − 4) = ( 4( − 1) ) = ( 4e

1 4

1 4

)

1 ( π + 2 kπ ) i 4

⇒ zk = 2 e

, k = 0, 1,2,3

 π kπ  + 4 2

 i 

, k = 0,1, 2, 3

For k=0

z0 = 2 e π π = 2 ( cos 4 + i sin 4 ) = 1+ i iπ 4

For k=1 π π   + i 4 2

z1 = 2 e  π π   π π  = 2 cos +  + i sin  +   4 2   4 2 1   1 = 2− +i  2 2  = −1 + i

For k=2 π   +π  i 4 

z2 = 2 e π    = 2 cos π +  + i sin ( π + π 4 )  4    1   1 = 2 − −  = −1 − i 2 2 

For k=3  π 3π  + 4 2

 i 

z3 = 2 e 7π 7π   = 2 cos + i sin  4 4  =1 − i

simple poles are z 0 = 1 + i & z1 = −1 + i which are inside C. 3 iaz

z e Re s f ( z ) = Re s 4 z =z z =1+i z +4 3 iaz ze = z =1+i 3 4z 1 ia ( 1+i ) = e 4 0

3 iaz

ze Re s f ( z ) = 3 z=z 4z

z = −1+ i

1

1 ia ( −1+i ) = e 4 1 − ai .e = e 4 −a

1 − a ai − ai Re s f ( z ) + Re s f ( z ) = e [ e + e ] z= z z= z 4 1 −a = e ( cos a + i sin a + cos a − i sin a ) 4 1 −a = e cos a 2 0

1

Now we have R

3 iax

x e dx + f ( z ) dz ∫ x 4 +4 ∫ -R CR =2πi ∑Re s f ( z ) 1 −a =2πi e cos a 2 −a =πie cos a

Taking Imaginary parts, we have R

3 iax

xe Im ∫ 4 dx + Im ∫ f ( z ) dz x + 4 -R C R

= Im(2πi ∑ Re s f ( z ) ) −a

= π e cos a

Hence R

3

x sin ax dx + Im f ( z ) dz ∫ x4 + 4 ∫ -R CR =π e

−a

cos a

We have 3 iaz

ze f(z) = 4 z +4 3

z Let φ ( z ) = 4 z +4

On C R : z = R , we get 3

R φ ( z) ≤ 4 → 0 as R → ∞ R −4 Hence by Jordan' s Lemma lim

R →∞

φ ( z ) e dz = 0 ∫ iaz

CR

Thus, on taking lim it when R → ∞, we get ∞

3

x sin ax −a dx = π e cos a ∫ x4 + 4 −∞

3

x sin ax −a ⇒ ∫ dx = π e cos a 4 −∞ x + 4 ∞

SEC 78 • Advice 3: • Page No. 280: Q. Nos. : 1 - 6

Definite integrals involving sines and cosines : Consider the integral 2π

I = ∫ F ( cos θ , sin θ ) dθ 0

iθ

Let z = e , 0 ≤ θ ≤ 2π dz iθ ⇒ dz = i.e dθ ⇒ = dθ iz

Also z = 1 1 1  1 iθ − iθ  z +  = ( e + e ) = cos θ 2 z 2 1 1  1 iθ − iθ  z −  = ( e − e ) = sin θ 2i  z  2i

2π

∴ I = ∫ F ( cos θ , sin θ ) dθ 0

= ∫ f ( z ) dz = 2πi ∑ Re s f ( z ) C : z =1

dθ Q1. Evaluate I = ∫ 0 5 + 4 sin θ 2π

∴ Let z = e

iθ

dz then I = ∫ 1 1  C : z =1  iz  5 + 4  z −   2i  z   dz =∫ 2 C 5iz + 2 z − 2

2 z + 5iz − 2 2 = 2 z + 4iz + iz − 2 = 2 z ( z + 2i ) + i ( z + 2i ) = ( z + 2i )( 2 z + i ) i  = 2( z + 2i )  z +  2  2

∴ I = ∫ f ( z ) dz , c

1 where f ( z ) = 2( z + 2i )( z + i 2 )

−i z = −2i, are simple poles of f(z) 2 -i but z = is the only pole which inside C. 2

1 Re s f ( z ) = z = −i 2 2( z + 2i )

z =−i / 2

1

1 = =  −i  3i 2 + 2i   2 

1 2π ∴ I = 2πi × = 3i 3

π

cos 2θ dθ Q.5 I = ∫ , a < 1 2 1 − 2 a cos θ + a 0

1 1 we have cosθ =  z + , 2 z 1 1 iθ sin θ =  z − , z = e 2i  z

∴1 − 2 a cosθ + a

2

1 1 2 = 1 − 2a  z +  + a 2 z a 1  = − ( z − a) z −  z a 

cos 2θ = 2 cos θ −1 2

2

1 1 = 2.  z +  −1 4 z

(

)

1 4 = 2 z +1 2z

π

∴I = ∫

cos 2θ . dθ 2

1 + a − 2 a cos θ 0

1 = 2

2π

cos 2θ . dθ

∫ 1 + a 2 − 2a cosθ

( z + 1)dz

0

1 =− ∫ 4ai c

4

1  z ( z − a ) z −  a  2

z +1 Let f ( z ) = 1  2 z ( z − a)  z −  a  1 then z = a, are simple poles a & z = 0 is a pole of order 2 of f(z) 4

1 ∴ a <1⇒ >1 a ∴ z = 0 & z = a are the only poles which are inside C.

z +1 Re s f ( z ) = z=a 1 2 z z−   a  z=a 4 a +1 = 2 a (a − 1) 4

    4 d z +1  Re s f ( z ) = z= 0 dz   1   ( z − a) z −    a   z= 0  a +1 = a 2

Re s f ( z ) +Re s f ( z ) z =a

z =0

a +1 a +1 = + 2 a (a −1) a 4

2

a +1 +a −1 = 2 a (a −1) 4

4

3

a = 2 a −1

3

1 a ∴I = − × 2πi × 2 4ai a −1 2 aπ = 2 1− a