More Integrals
The following integrals can be proved or worked out using complex analysis
methods. You may use the following equivalent known facts:
Z ∞
√
2
e−x /2 dx = 2π,
−∞
∞
Z
Z
2
e−πx dx = 1,
−∞
∞
2
e−x dx =
√
π.
−∞
1.
Z
C
1
2πi
dz =
(z − a)(z − b)
a−b
where C is the circle |z| = r (oriented counterclockwise) and |a| < r <
|b|.
2.
2
e−πξ =
Z
∞
2
e−πx e−2πixξ dx =
3.
∞
Z
0
Z
Z
2
sin(x ) dx =
0
0
5.
Z
0
6.
∞
−∞
7.
2
e−πx e2πixξ dx, ξ ∈ R
1 − cos x
π
dx = .
2
x
2
∞
Z
∞
−∞
−∞
4.
Z
Z
∞
∞
√
2π
cos(x ) dx =
.
4
2
π
sin x
dx = .
x
2

|t| < 1
 π
sin x ixt
π/2 |t| = 1
e dx =

x
0
|t| > 1
∞
a
,
+ b2
0
Z ∞
b
e−ax sin(bx) dx = 2
a + b2
0
e−ax cos(bx) dx =
where a > 0.
1
a2
8.
∞
Z
eax
π
dx =
, 0 < a < 1.
x
1+e
sin(πa)
−∞
9.
∞
Z
−∞
e−2πixξ
1
dx =
, ξ ∈ R.
cosh πx
cosh πξ
10.
Z
∞
e−2πixξ
−∞
sin πa
2 sinh 2πaξ
dx =
, 0 < a < 1, ξ ∈ R.
cosh πx + cos πa
sinh 2πξ
11.
∞
Z
−∞
12.
Z
∞
−∞
13.
Z
∞
−∞
14.
Z
∞
−∞
cos x
πe−a
dx
=
,a > 0
x2 + a2
a
x sin x
dx = πe−a , a > 0
x2 + a2
e−2πixξ
π
dx = (1 + 2π|ξ|)e−2π|ξ| , ξ ∈ R.
2
2
(1 + x )
2
1
1 · 3 · · · (2n − 1)
dx =
π, n positive integers.
(1 + x2 )n+1
2 · 4 · · · (2n)
15.
2π
Z
0
16.
Z
0
2π
1
2πa
dx = 2
a > 1.
(a + cos x)2
(a − 1)3/2
2π
1
dx = √
a > |b|, a, b ∈ R.
a + b cos x
a2 − b2
17.
1
Z
ln(sin πx) dx = − ln 2.
0
18.
Z
0
∞
π
ln x
dx =
ln a, a > 0.
2
+a
2a
x2
2
19.
2π
Z
ln |1 − aeit | dt = 0, |a| ≤ 1.
0
20.
∞
Z
ae−2πixξ
dx = πe−2πa|ξ| , a > 0, ξ ∈ R.
a2 + x2
−∞
21.
Z
∞
e−2πa|ξ| e2πiξx dξ =
π(a2
−∞
22.
∞
Z
0
23.
∞
Z
∞
Z
0
25.
Z
xa−1
π
dx =
, 0 < a < 1.
1+x
sin πa
xa−1
dx = π cot aπ, 0 < a < 1.
1−x
0
24.
1
π
dx =
, n = 2, 3, · · · .
1 + xn
n sin(π/n)
∞
π
xa
dx =
, 0 < a < 1.
1 + x2
2 cos(πa/2)
0
26.
Z
∞
ln(1 + x2 )
dx, 0 < a < 2.
xa+1
0
27.
a
, a > 0, x ∈ R.
+ x2 )
Z
0
∞
1
a
sinh ax
dx = tan , |a| < π.
sinh πx
2
2
28.
∞
Z
0
29.
Z
∞
0
30.
Z
∞
−∞
sin x
x
sin x
x
2
π
.
2
dx =
3π
.
8
3
sin x
x
3
dx =
2
eitx dx.
31. The Gamma function Γ(a), a > 0 is defined as the integral
Z ∞
xa−1 e−x dx.
0
(i) Prove that the integral converges. (ii) Prove that for a > 1, Γ(a) =
√
(a − 1)Γ(a − 1). (iii) Prove that Γ(1/2) = π.
32.
∞
Z
xa−1 eix dx = Γ(a)eiaπ/2 , 0 < a < 1.
0
33.
Z
∞
sin(x2 )
dx =
x2
∞
√
sin(x)
dx
=
2π.
x3/2
0
34.
Z
0
35.
2
cos(x ) cos(wx) dx =
0
36.
Z
2
0
37.
∞
Z
0
Z
0
39.
Z
∞
π
1
cos (w2 − π)
2
4
√
∞
sin(x ) cos(wx) dx =
38.
π
.
2
√
∞
Z
r
∞
π
1
cos (w2 + π)
2
4
sin t −st
1
e dt = tan−1 , s > 0
t
s
sin x −x
1
2
e cos(wx) dx = tan−1 2
x
2
w
sin(wx) tan−1
0
2a
sinh aw −aw
dx = π
e
,a > 0
x
w
40.
Z
0
∞
xn e−ax e−iwx dx =
n!
(a−iw)n+1 , n nonnegative integer
(a2 + w2 )n+1
4
41.
Z
∞
n!
<((a+iw)n+1 ), n nonnegative integer
(a2 + w2 )n+1
xn e−ax cos(wx) dx =
0
42.
Z
∞
xn e−ax sin(wx) dx =
(a2
0
43.
√
∞
Z
n!
=((a+iw)n+1 ), n nonnegative integer
+ w2 )n+1
−x2
e
cos(2bx) dx =
0
44.
Z
∞
−x2
e
sin(2bx) dx = e
0
−b2
π −b2
e
2
Z
b
2
et dt
0
Hints:
2
1. Problem 2. Consider integral of e−z over the rectangle with vertices
at −a, a,
a + iξ, −a + iξ, and let a → ∞. This yields
Z ∞
√
2
2
e−x e−2xwi dx = πe−w .
−∞
Then make substitution. (Another proof not using complex method
can be found in Ref [5] p. 138.)
iz
2. Problem 3. Integrate f (z) = 1−e
over an appropriate contour. Note
z2
that for y = =z > 0, |f (z)| ≤ K/|z|2 , and that f (z) = − zi + E(z),
where K > 0 is a constant and E(z) is an entire function.
2
3. Problem 4. Integrate eiz over the boundary (oriented counterclockwise) of the sector bounded by z = Reit , 0 ≤ t ≤ π/4, the nonnegative
x-axis, and the diagonal line in the first quadrant. Let R → ∞. (Use
the inequality: sin θ/θ ≥ 2/π for 0 ≤ θ ≤ π/2.)
iz
4. Problem 5. Integrate ez over indented semicircle of radius R. Use
Z π
Z π/2
Z π/2
−R sin t
−R sin t
e
dt = 2
e
dt ≤ 2
e−R2t/π dt.
0
0
0
and Theorem 1 in section 16.4 of Ref[4] to show that
Z ∞ ix
e
dx = πi.
−∞ x
5
(1)
5. Problem 6. Write the integral as
Z
1 ∞ eix(t+1) − eix(t−1)
dx
2i −∞
x
and use (1).
√
6. Problem 7. Let A = a2 + b2 and u = (a − ib)/A. Integrate e−Az over
the sector of the circle |z| = R cut out by the positive x-axis and the
ray z = ut, t ≥ 0. Let R → ∞.
7. Problem 8. Integrate f (z) = eaz /(1 + ez ) over the rectangle with
vertices at R, R + 2πi, −R + 2πi, −R. Note that f (z) has a simple pole
at z = πi with residue −eaπi . Note that
aR iat e e
eaR
1 + eR eit ≤ eR − 1
and
−aR iat e
e
e−aR
1 + e−R eit ≤ 1 − e−R
8. Problem 9. Integrate f (z) = e−2πizξ / cosh πz over the rectangle with
vertices at −R, R, R + 2i, −R + 2i. Note that f (z) has simple poles at
i/2, 3i/2. Note that | cosh(πR + πti)| ≥ eπR − e−πR .
9. Problem 10. Integrate f (z) = (e−2πizξ sin πa)/(cosh πz + cos πa) over
the rectangle with vertices at −R, R, R + 2i, −R + 2i. Note that f (z)
has simple poles at (1 + a)i, (1 − a)i in the rectangle.
10. Problem 12. Integrate f (z) = zeiz /(z 2 + a2 ). Use
|z 2
|z|
2
≤
2
+a |
R
if |z| ≥ R for sufficiently large R. Argue as in Problem 5.
11. Problem 13. If ξ > 0, integrate over a semicircle below the x-axis
clockwise; if ξ < 0 integrate over a semicircle above the x-axis counterclockwise. The residue of
e−2πizξ
(1 + z 2 )2
at z = −i is
−
1 + 2πξ −2πξ
e
4i
6
12. Problem 17. Note that 1 − e2πiz = −2ieiπz sin πz. f (z) = Ln (1 −
e2πiz ) is analytic on the complement of the set {x + iy : x = n, y ≤
0, n positive integers }. Integrate f (z) on the rectangle with vertices
at Ri, 1 + Ri and indented quarter-circle corners at 0 and 1. Since
|1 − e2πiz |/|z| → 2π as z → 0 and limξ→0 ξ ln ξ = 0 the integral over
the quarter-circle at 0 tends to 0. Since e2πi(z−1) = e2πiz , the integral
over the quarter-circle at 1 also tends to 0. Let R → ∞ and the corners
shrink to 0. Then
Z 1
Ln (1 − e2πix ) dx = 0.
0
Note that Arg (ab) = Arg (a)+ Arg (b) if −π < Arg (a)+ Arg (b) ≤
π, so
Ln (−2ieiπx sin πx) = ln 2 + ln(sin πx) + i(−π/2 + πx).
z
13. Problem 18. Integrate f (z) = zLn
2 +a2 over the boundary S\A\B, where
S = {(x, y) : x2 + y 2 ≤ R2 , y ≥ 0}, A = {(x, y) : x2 + y 2 < 2 },
and B = {(x, y) : x < 0, y < η}, where > 0 and 0 < η < . Let
R → ∞, → 0.
14. Problem 19. For |a| < 1, Ln (1 − aeiz ) is analytic on the upper plane
y ≥ 0. For a = −1, 1, see the proof of Problem 17. Recall the real
part of Ln z is ln |z|.
15. Problem 21. Use symmetry to write the integral as one over (0, ∞).
Integrate a function of the form ef (z) on the boundary of {(x, y) : x ≥
0, 0 ≤ y ≤ i}. Take the real part of the result.
16. Problems 22,23. Integrate a semicircle with a neighborhood of −1, a
neighborhood of 0, and a neighborhood of of the positive x-axis removed. Or, for 22 only, the boundary of a full circular disk with a
neighborhood of 0 and a neighborhood of of the positive x-axis removed. (Use the branch of ln such that the 0 ≤ argz < 2π for all
z.)
17. Problem 24. Integrate over a circular sector with angle π/n.
18. Problem 25. Use the contour for Problem 18.
19. Problem 26. Try integration by parts, and use Problem 25.
7
az
e
20. Problem 27. Integrate sinh
πz over the boundary of {(x, y) : |x| ≤
R, 0 ≤ y ≤ 1} with neighborhoods of 0, i removed.
2iz
21. Problem 28. Integrate f (z) = e z 2−1 over an indented semicircle. Note
that f (z) has a simple pole at 0.
22. Problem 29. (−8i) sin3 x = ei3x − 3eix + 3e−ix − e−i3x . So
(−8i) sin3 x
ei3x − 3eix + 2 e−i3x − 3e−ix + 2
=
−
.
x3
x3
x3
By change of variable u = −x, we get
Z ∞ i3x
Z ∞
e − 3eix + 2
sin3 x
dx
=
2
dx.
(−8i)
3
x3
−∞
−∞ x
3iz
iz
+2
Now integrate f (z) = e −3e
over an indented semicircle. Note
z3
that f (z) has a simple pole at 0.
23. Problem 30. Rewrite f (z) = (sin2 z/z 2 )eitz as
2eitz − e(t+2)iz − e(t−2)iz
.
4z 2
Consider cases |t| > 2, −2 < t < 0 and 0 < t < 2 separately. Note that
0 is a removable singularity of f . If t > 2(t < −2), then integrate f (z)
over semicircle in the upper(lower) plane. If 0 < t < 2 then rewrite
f (z) as
2eitz − e(t+2)iz − 1 1 − e(t−2)iz
+
4z 2
4z 2
and integrate the first(second) summand over indented semicircle in
the upper(lower) plane. Similar for −2 < t < 0. Note that the above
mentioned summands have simple pole at 0.
24. Problem 31. For (ii), use integration by parts. For part (iii), substitute
√
u = x.
25. Problem 32. Contour integrate z a−1 eiz over the boundary of the sector
{z : z = reiθ , 0 ≤ r ≤ R, 0 ≤ θ ≤ π/2}, dented at the origin. Let
R → ∞.
26. Problem 33. Integration by parts with dv =
4.
8
1
x2
dx, then use Problem
27. Problem 34. Substitute u =
√
x and use Problem 33.
R∞
28. Problems 35 (Problem 36 is similar). Write the integral as 12 −∞ cos(x2 )e−iwx dx,
R∞
2
2
then as 41 −∞ (eix + e−ix )e−iwx dx. Complete the squares and make
substitutions u = x ± w2 . Then use Problem 4.
29. Problem 37. This is just the restatement that the Laplace transform
of sin t/t is tan−1 (1/s). To prove directly, rewrite the integral as
Z ∞Z ∞
e−ut sin t du dt.
0
Note that
Z ∞Z ∞
e
0
−ut
s
Z
∞
| sin t| du dt =
0
s
| sin t| −st
e dt ≤
t
Z
0
∞
e−st =
1
<∞
s
By Ref [6] Theorem 8.8 (a), p. 164, e−ut | sin t| and hence e−ut sin t
is (Lebesgue) integrable, and by (c), the order of integration can be
interchanged. Since the Laplace transform of sin t is 1/(s2 + 1), the
result follows. (Note that π/2 − tan−1 x = cot−1 x = tan−1 (1/x) if
x > 0.)
30. Problem 38. Use trig identity and Problem 37. Note that tan−1 (1 +
w) + tan−1 (1 − w) = tan−1 (2/w2 ).
−1 x . Then use integration by
31. Problem 39. Rewrite tan−1 2a
x as cot
2a
x
−1
parts with u = cot 2a and dv = sin(wx) dx. Use Problem 11.
32. Problem 40. This is just the Laplace transform of xn at s = a + iw.
33. Problems 41, 42. Note that <((a + iw)n+1 ) = <((a − iw)n+1 ) and
=((a + iw)n+1 ) = −=((a − iw)n+1 ). This is clear when a + iw is written in its polar form.
2
34. Problem 43. Integrate e−z around the boundary of the rectangle with
vertices −a, a, a + ib, −a + ib, letting a → ∞. Or use Problem 2, with
2
some scale change; note that e−x cos(2bx) is even.
2
35. Problem 44. Integrate e−z around the boundary of the rectangle with
vertices −ib, a − ib, a + ib, ib, letting a → ∞.
9
References
1. Shakarchi and Stein, Complex Analysis.
2. Alan Jeffrey, Complex analysis and applications.
3. Ahlfors, Complex analysis.
4. Kreyszig, Advanced Engineering Mathematics (9th or 10th edition).
5. Shakarchi and Stein, Fourier Analysis.
6. Rudin, Real and Complex Analysis (3rd edition).
10