The Frullani integrals
Notes by G.J.O. Jameson
We consider integrals of the form
∞
Z
If (a, b) =
0
f (ax) − f (bx)
dx,
x
where f is a continuous function (real or complex) on (0, ∞) and a, b > 0. If f (x) tends to
a non-zero limit at 0, then the separate integrals of f (ax)/x and f (bx)/x diverge at 0, and a
similar comment applies at infinity. The point is that under suitable conditions, the integral
of the difference converges.
The basic theorem is as follows. I do not know many books or articles where proofs
are given: two references (for which I am grateful to Nick Lord) are [Fer, p. 134–135] and
[Tr].
FRU1 THEOREM. Consider the following conditions:
(C1)
f (x) → c0 as x → 0+ ;
(C2)
f (x) → c∞ as x → ∞;
(C3)
there exists K such that |F (x)| ≤ K for all x > 0, where F (x) =
Rx
0
f (t) dt.
Under conditions (C1) and (C2), we have
If (a, b) = (c0 − c∞ )(log b − log a).
(1)
Under conditions (C1) and (C3), we have
If (a, b) = c0 (log b − log a).
(2)
Proof. We may assume that b > a: interchanging a and b then gives the case b < a.
Let 0 < δ < X. The substitution ax = y gives
Z X
Z aX
f (ax)
f (y)
dx =
dy.
x
y
δ
aδ
So
Z
δ
X
Z bX
f (y)
f (y)
dy −
dy
y
y
aδ
bδ
Z bδ
Z bX
f (y)
f (y)
=
dy −
dy
y
y
aδ
aX
= I(δ) − I(X),
f (ax) − f (bx)
dx =
x
Z
1
aX
where we write
br
Z
f (y)
dy
y
I(r) =
ar
for any r > 0. Let ε > 0 be given. Under condition (C1), there exists δ0 > 0 such that if
0 < y ≤ bδ0 , then |f (y) − c0 | ≤ ε. Then for δ ≤ δ0 , we have I(δ) = c0 (log b − log a) + r1 (δ),
where
Z
bδ
f (y) − c0
dy,
y
r1 (δ) =
aδ
hence
Z
bδ
|r1 (δ)| ≤
aδ
ε
dy = ε(log b − log a).
y
+
So I(δ) → c0 (log b − log a) as δ → 0 .
In exactly the same way, condition (C2) implies that I(X) → c∞ (log b − log a) as
X → ∞.
Under condition (C3), integration by parts gives
h F (y) ibX Z bX F (y)
I(X) =
+
dy,
y aX
y2
aX
hence
2K
|I(X)| ≤
+K
aX
Z
bX
aX
1
3K
dy <
,
2
y
aX
so I(X) → 0 as X → ∞. (Of course, this also shows that
R∞
1
f (x)
x
dx converges; this
statement could be taken as the hypothesis instead of (C3)).
In particular, If (1, b) equals (c0 − c∞ ) log b in case (1), and c0 log b in case (2).
We record a number of particular examples which are transparently cases of (1) or (2),
without repeatedly writing out the integral expressions:
f (x)
If (a, b)
e−x
1/(1 + x2 )
cos x
−x
e cos x
sin x/x
tan−1 x
tanh x
log b − log a
log b − log a
log b − log a
log b − log a
log b − log a
π
(log a − log b)
2
log a − log b
Among these examples, cos x is the only one satisfying (C3), but not (C2). A simple
alternative proof for this case is given in [Jam, p. 280].
2
Many further examples of Frullani integrals are given in [AABM].
We now state a simple extension of the Theorem.
Pn
FRU2. Let G(x) =
j=1
mj f (aj x), where aj > 0 for 1 ≤ j ≤ n and
Pn
j=1
mj = 0. If
f satisfies (C1) and (C2), then
Z ∞
n
X
G(x)
dx = (c∞ − c0 )
mj log aj .
x
0
j=1
(3)
If f satisifes (C1) and (C3), the same applies with c∞ replaced by 0.
Proof. Write Mj = m1 + · · · + mj . By Abel summation, since Mn = 0,
G(x) =
n−1
X
Mj [f (aj x) − f (aj+1 x)].
j=1
By (1),
Z
∞
0
n−1
n
X
X
G(x)
dx = (c∞ − c0 )
Mj (log aj − log aj+1 ) = (c∞ − c0 )
mj log aj .
x
j=1
j=1
Double integral method for (1).
For monotonic f , the following is an alternative method for (1) (but not (2)). The
method has been in circulation for a long time, at least for the special case f (x) = e−x . For
example, it can be seen in [Cou, p. 240].
Note that for x, y > 0,
1 d
1 d
f (xy) =
f (xy) = f 0 (xy).
y dx
x dy
Hence
Z
b
0
f (xy) dy =
h1
x
a
ib
f (xy)
=
y=a
f (bx) − f (ax)
.
x
0
Since f (x) is of constant sign, reversal of the following double integral is justfied:
Z ∞Z b
Z bZ ∞
0
−If (a, b) =
f (xy) dy dx =
f 0 (xy) dx dy.
0
But
Z
a
a
∞
0
f (xy) dx =
0
So
Z
−If (a, b) = (c∞ − c0 )
a
b
h1
y
f (xy)
i∞
x=0
0
1
= (c∞ − c0 ).
y
1
dy = (c∞ − c0 )(log b − log a).
y
3
Some applications of the case f (x) = e−x .
This is undoubtedly the best known Frullani integral. Written out explicitly, the statement is
∞
e−ax − e−bx
dx = log b − log a.
x
0
The substitution x = − log y transforms (4) to
Z 1 b−1
y
− y a−1
dy = log b − log a.
log y
0
Z
(4)
So, for example,
1
Z
0
y−1
dy = log 2
log y
(cf. [BM, p. 97]).
Now consider the integral
Z
I1 =
0
∞
1
(1 − e−ax )(1 − e−bx ) dx.
x2
Integration by parts gives I1 = A1 + J1 , where
h 1
i∞
−ax
−bx
A1 = − (1 − e )(1 − e ) ,
x
0
Z ∞
1
J1 =
ae−ax + be−bx − (a + b)e−(a+b)x .
x
0
−x
Since 0 < 1 − e < x for x > 0, we have x1 (1 − e−ax )(1 − e−bx ) < abx, hence A1 = 0. So by
(3),
I1 = J1 = (a + b) log(a + b) − a log a − b log b.
Next, consider the function
Z
E(x) =
x
∞
e−t
dt.
t
By reversal of the implied double integral, one sees easily that
Z ∞
I2 =
e−ax E(x) dx,
R∞
0
E(x) dx = 1. Now let
0
where a > 0. Reversing the double integral and applying (4), we find
Z ∞
Z ∞ −t
e
−ax
I2 =
e
dt dx
t
0
x
Z ∞ −t Z t
e
=
e−ax dx dt
t
Z0 ∞ −t 0 −at
e (1 − e )
=
dt
at
0
1
=
log(1 + a).
a
4
Applications of the case f (x) = cos x
We describe two applications of this case. First, since 2 sin ax sin bx = cos(a − b)x −
cos(a + b)x, we have, for a > b > 0,
Z ∞
a+b
sin ax sin bx
dx = 12 log
.
x
a−b
0
Second, consider the integral
Z
I3 =
0
∞
sin5 x
dx.
x2
Since 16 sin5 x = 10 sin x − 5 sin 3x + sin 5x, we have by (3)
h sin5 x i∞ Z ∞ 1
+
(10 cos x − 15 cos 3x + 5 cos 5x) dx
16I3 =
−
x 0
x
0
= −(10 log 1 − 15 log 3 + 5 log 5)
= 15 log 3 − 5 log 5.
Integrals of this type are discussed more generally in [Tr].
References
[AABM]
Matthew Albano, Tewodros Amdeberhan, Erin Beyerstedt and Victor H. Moll,
The integrals in Gradshteyn and Ryzhik. Part 15: Frullani integrals,
Scientica, Series A: Mathematical Sciences 19 (2010), 113–119.
[BM]
George Boros and Victor H. Moll, Irresistible Integrals, Cambridge Univ. Press
(2004).
[Cou]
R. Courant, Differential and Integral Calculus, vol. II, Blackie & Son (1936).
[Fer]
[Jam]
[Tr]
W. L. Ferrar, Integral Calculus, Oxford Univ. Press (1958).
G. J. O. Jameson, Sine, cosine and exponential integrals, Math. Gazette 99
(2015), 276–289.
J. Trainin, Integrating expressions of the form
94 (2010), 216–223.
sinn x
xm
and others, Math. Gazette
updated 16 November 2015
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