π
until it is inside
ln
z
and
z
3
/
4
I
3
=
0
ˆ
∞
ln (
xe
i
2
π
)
(
xe
i
2
π
)
3
/
4
(1 +
x
)
dx
=

i
I
+
∞
ˆ
0
2
πi
x
3
/
4
(1 +
x
)
dx
(7)
where I used
e

i
3
π/
2
=
i
, and where the minus sign in the 2nd line arises from flipping the
integration limits from the first line. Note that a new integral has popped up on the right
handside above, along with the integral that we want to evaluate. So it looks like we are
stuck with evaluating the extra integral
I
extra
=
∞
ˆ
0
1
x
3
/
4
(1 +
x
)
dx
7
before we can recover the one we set out to evaluate.
But, before we get too upset with
having an additional task to perform, let’s substitute Eqs. (6) and (7) into Eqs. (4) and (5),
to obtain
I
C
= (1

i
)
×
I
+ 2
πI
extra
=
√
2
π
2
(1 +
i
)
(8)
Now
here
comes
a
very
clever
idea,
due
to
Amar
Masalmeh: we can simply take the
imaginary part of the two sides of the above equation, which eliminates the term with the
extra integral, and allows us to immediately read off the value of the integral that we want!
We thereby obtain the desired result
I
=

√
2
π
2
At this point, one could also deduce the extra integral, but taking the
real part of Eq. (8),
and using the result we just obtained for
I
, to find
I
extra
=
√
2
π
It turns out that we already evaluated an integral of the form of
I
extra
, in class, which
provides a crosscheck of the above
∞
ˆ
0
x
p

1
1 +
x
dx
=
π
sin
πp
0
< p <
1
Setting
p
= 1
/
4
, we recover our previous result for
I
extra
!
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 Winter '20
 Physics, Unit Circle, Methods of contour integration, Complex Plane, Pole, Branch point