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1 Mathematical Physics
Residue at exp(πi/4) = lim z→exp(
πi
4 )
z − exp
πi
4
1
z 4 + 1
=
1
4z 3 =
1
4
exp
−
3πi
4
Residue at exp(3πi/4) = lim z→exp(
3πi
4 )
z − exp
3πi
4
1
z 4 + 1
=
1
4z 3 =
1
4
exp
−
3πi
4
Thus
c
dz
z 4 + 1
= 2πi
1
4
exp
−
3πi
4
+
1
4
exp
−
3πi
4
=
π
√
2
Thus
R
−R
dx
x 4 + 1
+
dz
z 4 + 1
=
π
√
2
Taking the limit of both sides as R → ∞
lim R→∞
+R
−R
dx
x 4 + 1
=
∞
−∞
dx
x 4 + 1
=
π
√
2
It follows that
∞
0
dx
x 4 + 1
=
π
2
√
2
Fig. 1.17 Closed contour
consisting of line from −R
to R and the semi-circle Γ
1.3.12 Calculus of Variation
1.89 Let I =
x1
x0
F(x, y, y
)dx
(1)
Here I =
x 1
x 0
1 +
dy
dx
2
dx
(2)
Now the Euler equation is
∂ F
∂ y
−
d
dx
∂ F
∂ y
= 0
( 3 )