11.5. THE METHOD OF RESIDUES 267
= limz→ 1
2√
2+ 12 i√
2
(ln |z|+ iarg1 (z))+(
z−(
12
√2+ 1
2 i√
2))
(1/z)
4z3
=
ln(√
12 +
12
)+ i π
4
4(
12
√2+ 1
2 i√
2)3 =
(1
32− 1
32i)√
2π
Similarly
res(
f ,−12
√2+
12
i√
2)=
ln(√
12 +
12
)+ i 3π
4
4(− 1
2
√2+ 1
2 i√
2)3 =
332
√2π +
332
i√
2π.
Of course it is necessary to consider the integral along the small semicircle of radius r. Thisreduces to ∫ 0
π
ln |r|+ it
1+(reit)4
(rieit)dt
which clearly converges to zero as r→ 0 because r lnr→ 0. Therefore, taking the limit asr→ 0, ∫
large semicircle
L(z)1+ z4 dz+ lim
r→0+
∫ −r
−R
ln(−t)+ iπ1+ t4 dt+
limr→0+
∫ R
r
ln t1+ t4 dt = 2πi
(332
√2π +
332
i√
2π +1
32
√2π− 1
32i√
2π
).
Observing that∫
large semicircleL(z)1+z4 dz→ 0 as R→ ∞,
e(R)+2 limr→0+
∫ R
r
ln t1+ t4 dt + iπ
∫ 0
−∞
11+ t4 dt =
(−1
8+
14
i)
π2√
2
where e(R)→ 0 as R→ ∞. This becomes
e(R)+2 limr→0+
∫ R
r
ln t1+ t4 dt + iπ
(√2
4π
)=
(−1
8+
14
i)
π2√
2.
Now letting r→ 0+ and R→ ∞,
2∫
∞
0
ln t1+ t4 dt =
(−1
8+
14
i)
π2√
2− iπ
(√2
4π
)=−1
8
√2π
2,
and so∫
∞
0ln t
1+t4 dt = − 116
√2π2, which is probably not the first thing you would thing of.
You might try to imagine how this could be obtained using only real analysis. I don’t haveany idea how to get this one from standard methods of calculus. Perhaps some sort ofpartial fractions might do the job but even if so, it would be very involved.