1000 Solved Problems in Modern Physics

74 1 Mathematical Physics

Residue at exp(πi/4)=limz→exp(π 4 i)

{

z−exp

(

πi

4

)

1

z^4 + 1

}

=

1

4 z^3

=

1

4

exp

(

−

3 πi

4

)

Residue at exp(3πi/4)=limz→exp( 3 π 4 i)

{

z−exp

(

3 πi

4

)

1

z^4 + 1

}

=

1

4 z^3

=

1

4

exp

(

−

3 πi

4

)

Thus

∮

c

dz

z^4 + 1

= 2 πi

{

1

4

exp

(

−

3 πi

4

)

+

1

4

exp

(

−

3 πi

4

)}

=

π

√

2

Thus

∫ R

−R

dx

x^4 + 1

+

∫

dz

z^4 + 1

=

π

√

2

Taking the limit of both sides asR→∞

limR→∞

∫+R

−R

dx

x^4 + 1

=

∫∞

−∞

dx

x^4 + 1

=

π

√

2

It follows that

∫∞

0

dx

x^4 + 1

=

π

2

√

2

Fig. 1.17Closed contour
consisting of line from−R
to R and the semi-circleΓ

1.3.12 CalculusofVariation...............................

1.89LetI=

∫x 1

x 0

F(x,y,y′)dx (1)

HereI=

∫x 1

x 0

√

1 +

(

dy

dx

) 2

dx (2)

Now the Euler equation is

∂F

∂y

−

d

dx

(

∂F

∂y′

)

=0(3)