Les integrals de Frullani són un tipus específic d'integral impròpia que rep el nom del matemàtic italià Giuliano Frullani , qui les va esmentar per primera vegada en una carta el 1821 i publicada el 1828.
Les integrals són de la forma
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
{\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x}
on
f
(
x
)
{\displaystyle {f(x)}}
x
≥
0
{\displaystyle {x\geq 0}}
f
(
x
)
{\displaystyle {f(x)}}
∞
{\displaystyle {\infty }}
La següent fórmula per a la seva solució general es compleix en determinades condicions:
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
=
(
f
(
∞
)
−
f
(
0
)
)
ln
a
b
{\displaystyle {\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x={\Big (}f(\infty )-f(0){\Big )}\ln {\frac {a}{b}}}}
Una demostració simple de la fórmula es pot arribar expandint l'integrand en una integral, i després utilitzant el teorema de Fubini per intercanviar les dues integrals:
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
=
∫
0
∞
[
f
(
x
t
)
x
]
t
=
b
t
=
a
d
x
=
∫
0
∞
∫
b
a
f
′
(
x
t
)
d
t
d
x
=
∫
b
a
∫
0
∞
f
′
(
x
t
)
d
x
d
t
=
∫
b
a
[
f
(
x
t
)
t
]
x
=
0
x
=
∞
d
t
=
∫
b
a
f
(
∞
)
−
f
(
0
)
t
d
t
=
(
f
(
∞
)
−
f
(
0
)
)
(
ln
(
a
)
−
ln
(
b
)
)
=
(
f
(
∞
)
−
f
(
0
)
)
ln
(
a
b
)
{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,dx&=\int _{0}^{\infty }\left[{\frac {f(xt)}{x}}\right]_{t=b}^{t=a}\,dx\\&=\int _{0}^{\infty }\int _{b}^{a}f'(xt)\,dt\,dx\\&=\int _{b}^{a}\int _{0}^{\infty }f'(xt)\,dx\,dt\\&=\int _{b}^{a}\left[{\frac {f(xt)}{t}}\right]_{x=0}^{x=\infty }\,dt\\&=\int _{b}^{a}{\frac {f(\infty )-f(0)}{t}}\,dt\\&={\Big (}f(\infty )-f(0){\Big )}{\Big (}\ln(a)-\ln(b){\Big )}\\&={\Big (}f(\infty )-f(0){\Big )}\ln {\Big (}{\frac {a}{b}}{\Big )}\\\end{aligned}}}
Tingueu en compte que la integral de la segona línia anterior s'ha pres sobre l'interval
[
b
,
a
]
{\displaystyle [b,a]}
[
a
,
b
]
{\displaystyle [a,b]}
Si
f
(
x
)
∈
C
[
0
,
+
∞
)
{\displaystyle f(x)\in C[0,+\infty )\ }
∀
A
>
0
∃
∫
A
∞
f
(
x
)
x
d
x
{\displaystyle \ \forall A>0\ \exists \int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx}
∫
0
∞
f
(
α
x
)
−
f
(
β
x
)
x
d
x
=
f
(
0
)
ln
(
β
α
)
(
α
>
0
,
β
>
0
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=f(0)\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}\ \ (\alpha >0,\beta >0)\ }
Demostració:
lim
A
→
+
0
(
∫
A
∞
f
(
α
x
)
−
f
(
β
x
)
x
d
x
)
=
lim
A
→
+
0
(
∫
A
∞
f
(
α
x
)
x
d
x
−
∫
A
∞
f
(
β
x
)
x
d
x
)
=
{\displaystyle \lim \limits _{A\to +0}{\Biggl (}{\int \limits _{A}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx}{\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}{{\int \limits _{A}^{\infty }{\frac {f(\alpha x)}{x}}\,dx}-{\int \limits _{A}^{\infty }{\frac {f(\beta x)}{x}}\,dx}}{\Biggr )}=}
=
{
∀
A
>
0
∃
∫
A
∞
f
(
x
)
x
d
x
⇒
∫
A
∞
f
(
x
)
x
d
x
=
F
(
∞
)
−
F
(
A
)
⇒
∫
A
∞
f
(
α
x
)
x
d
x
=
∫
α
A
∞
f
(
x
)
x
d
x
=
F
(
∞
)
−
F
(
α
A
)
}
{\displaystyle =\left\{\forall A>0\ \exists \int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx\ \Rightarrow {\int \limits _{A}^{\infty }{\frac {f(x)}{x}}\,dx=F(\infty )-F(A)}\Rightarrow {\int \limits _{A}^{\infty }{\frac {f(\alpha x)}{x}}\,dx=\int \limits _{\alpha A}^{\infty }{\frac {f(x)}{x}}\,dx}=F(\infty )-F(\alpha A)\right\}}
=
{\displaystyle =}
=
lim
A
→
+
0
(
F
(
∞
)
−
F
(
α
A
)
−
F
(
∞
)
+
F
(
β
A
)
)
=
lim
A
→
+
0
(
F
(
β
A
)
−
F
(
α
A
)
)
=
lim
A
→
+
0
(
∫
α
β
f
(
A
x
)
x
d
x
)
=
{\displaystyle =\lim \limits _{A\to +0}{\Biggl (}F(\infty )-F(\alpha A)-F(\infty )+F(\beta A){\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}F(\beta A)-F(\alpha A){\Biggr )}=\lim \limits _{A\to +0}{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(Ax)}{x}}\,dx{\Biggr )}=}
=
lim
A
→
+
0
(
f
(
A
ξ
)
∫
α
β
1
x
d
x
)
{\displaystyle =\lim \limits _{A\to +0}{\Biggl (}f(A\xi )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}}
=
lim
A
→
+
0
(
f
(
A
ξ
)
(
ln
(
β
)
−
ln
(
α
)
)
)
{\displaystyle =\lim \limits _{A\to +0}{\Biggl (}f(A\xi ){\biggl (}\ln(\beta )-\ln(\alpha ){\biggr )}{\Biggr )}}
=
lim
A
→
+
0
(
f
(
A
ξ
)
)
ln
(
β
α
)
=
{\displaystyle =\lim \limits _{A\to +0}{\biggl (}f(A\xi ){\biggr )}\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}=}
=
{
ξ
∈
[
α
,
β
]
⇒
lim
A
→
+
0
A
ξ
=
0
,
f
(
x
)
∈
C
[
0
,
+
∞
)
⇒
lim
A
→
+
0
f
(
A
ξ
)
=
f
(
0
)
}
=
f
(
0
)
ln
(
β
α
)
.
{\displaystyle =\left\{\xi \in [\alpha ,\beta ]\Rightarrow \lim \limits _{A\to +0}{A\xi }=0,f(x)\in C[0,+\infty )\Rightarrow \lim \limits _{A\to +0}{f(A\xi )=f(0)}\right\}=f(0)\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}.}
Si
f
(
x
)
∈
C
[
0
,
+
∞
)
{\displaystyle f(x)\in C[0,+\infty )}
∃
lim
x
→
+
∞
f
(
x
)
<
+
∞
{\displaystyle \exists \lim \limits _{x\to +\infty }f(x)<+\infty \ }
∫
0
∞
f
(
α
x
)
−
f
(
β
x
)
x
d
x
=
(
f
(
0
)
−
f
(
+
∞
)
)
ln
(
β
α
)
(
α
>
0
,
β
>
0
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=(f(0)-f(+\infty ))\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}\ \ (\alpha >0,\beta >0)}
Demostració:
∫
0
∞
f
(
α
x
)
−
f
(
β
x
)
x
d
x
=
lim
ϵ
→
0
,
Δ
→
∞
(
∫
ϵ
A
f
(
α
x
)
−
f
(
β
x
)
x
d
x
+
∫
A
Δ
f
(
α
x
)
−
f
(
β
x
)
x
d
x
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=\lim \limits _{\epsilon \to 0,\Delta \to \infty }{\Biggl (}\int \limits _{\epsilon }^{A}{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx+\int \limits _{A}^{\Delta }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx{\Biggr )}}
=
{\displaystyle =}
=
{
ρ
(
ϵ
,
A
)
<
∞
,
f
(
x
)
x
∈
C
[
ϵ
,
A
]
⇒
∫
ϵ
A
f
(
x
)
x
d
x
=
F
(
A
)
−
F
(
ϵ
)
⇒
∫
ϵ
A
f
(
α
x
)
x
d
x
=
F
(
α
A
)
−
F
(
α
ϵ
)
}
{\displaystyle =\left\{\rho {\bigl (}\epsilon ,A{\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[\epsilon ,A]\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(x)}{x}}\,dx=F(A)-F(\epsilon )\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(\alpha x)}{x}}\,dx=F(\alpha A)-F(\alpha \epsilon )\right\}}
=
{\displaystyle =}
=
lim
ϵ
→
0
,
Δ
→
+
∞
(
F
(
α
A
)
−
F
(
α
ϵ
)
−
F
(
β
A
)
+
F
(
β
ϵ
)
+
∫
A
Δ
f
(
α
x
)
−
f
(
β
x
)
x
d
x
)
=
{\displaystyle =\lim \limits _{\epsilon \to 0,\Delta \to +\infty }{\Biggl (}F(\alpha A)-F(\alpha \epsilon )-F(\beta A)+F(\beta \epsilon )+\int \limits _{A}^{\Delta }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx{\Biggr )}=}
=
{
ρ
(
A
,
Δ
)
<
∞
,
f
(
x
)
x
∈
C
[
A
,
Δ
]
⇒
∫
A
Δ
f
(
x
)
x
d
x
=
F
(
Δ
)
−
F
(
A
)
⇒
∫
A
Δ
f
(
α
x
)
x
d
x
=
F
(
α
Δ
)
−
F
(
α
A
)
}
=
{\displaystyle =\left\{\rho {\bigl (}A,\Delta {\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[A,\Delta ]\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(x)}{x}}\,dx=F(\Delta )-F(A)\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(\alpha x)}{x}}\,dx=F(\alpha \Delta )-F(\alpha A)\right\}=}
=
lim
ϵ
→
+
0
,
Δ
→
+
∞
(
F
(
α
A
)
−
F
(
α
ϵ
)
−
F
(
β
A
)
+
F
(
β
ϵ
)
+
F
(
α
Δ
)
−
F
(
α
A
)
−
F
(
β
Δ
)
+
F
(
β
A
)
)
=
{\displaystyle =\lim \limits _{\epsilon \to +0,\Delta \to +\infty }{\Biggl (}F(\alpha A)-F(\alpha \epsilon )-F(\beta A)+F(\beta \epsilon )+F(\alpha \Delta )-F(\alpha A)-F(\beta \Delta )+F(\beta A){\Biggr )}=}
=
lim
ϵ
→
+
0
(
F
(
β
ϵ
)
−
F
(
α
ϵ
)
)
−
lim
Δ
→
+
∞
(
F
(
β
Δ
)
−
F
(
α
Δ
)
)
=
lim
ϵ
→
+
0
(
∫
α
β
f
(
ϵ
x
)
x
d
x
)
−
lim
Δ
→
+
∞
(
∫
α
β
f
(
Δ
x
)
x
d
x
)
=
{\displaystyle =\lim \limits _{\epsilon \to +0}{\biggl (}F(\beta \epsilon )-F(\alpha \epsilon ){\biggr )}-\lim \limits _{\Delta \to +\infty }{\biggl (}F(\beta \Delta )-F(\alpha \Delta ){\biggr )}=\lim \limits _{\epsilon \to +0}{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(\epsilon x)}{x}}\,dx{\Biggr )}-\lim \limits _{\Delta \to +\infty }{\Biggl (}\int \limits _{\alpha }^{\beta }{\frac {f(\Delta x)}{x}}\,dx{\Biggr )}=}
=
lim
ϵ
→
+
0
(
f
(
ϵ
η
)
∫
α
β
1
x
d
x
)
−
lim
Δ
→
+
∞
(
f
(
Δ
μ
)
∫
α
β
1
x
d
x
)
{\displaystyle =\lim \limits _{\epsilon \to +0}{\Biggl (}f(\epsilon \eta )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}-\lim \limits _{\Delta \to +\infty }{\Biggl (}f(\Delta \mu )\int \limits _{\alpha }^{\beta }{\frac {1}{x}}\,dx{\Biggr )}}
=
(
lim
ϵ
→
+
0
f
(
ϵ
η
)
−
lim
Δ
→
+
∞
f
(
Δ
μ
)
)
(
ln
(
β
)
−
ln
(
α
)
)
{\displaystyle ={\biggl (}\lim \limits _{\epsilon \to +0}f(\epsilon \eta )-\lim \limits _{\Delta \to +\infty }f(\Delta \mu ){\biggr )}{\biggl (}\ln(\beta )-\ln(\alpha ){\biggr )}}
=
{\displaystyle =}
=
{
η
,
μ
∈
[
α
,
β
]
⇒
lim
ϵ
→
+
0
ϵ
η
=
0
,
lim
Δ
→
+
∞
Δ
μ
=
+
∞
,
f
(
x
)
∈
C
[
0
,
+
∞
]
⇒
lim
ϵ
→
+
0
f
(
ϵ
η
)
=
f
(
0
)
,
lim
Δ
→
+
∞
f
(
Δ
μ
)
=
f
(
+
∞
)
}
=
{\displaystyle =\left\{\eta ,\mu \in [\alpha ,\beta ]\Rightarrow \lim \limits _{\epsilon \to +0}\epsilon \eta =0,\lim \limits _{\Delta \to +\infty }\Delta \mu =+\infty ,f(x)\in C[0,+\infty ]\Rightarrow \lim \limits _{\epsilon \to +0}f(\epsilon \eta )=f(0),\lim \limits _{\Delta \to +\infty }f(\Delta \mu )=f(+\infty )\right\}=}
=
(
f
(
0
)
−
f
(
+
∞
)
)
ln
(
β
α
)
.
{\displaystyle ={\biggl (}f(0)-f(+\infty ){\biggr )}\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}.}
Si
f
(
x
)
∈
C
(
0
,
+
∞
)
{\displaystyle f(x)\in C(0,+\infty )\ }
∀
A
>
0
∃
∫
0
A
f
(
x
)
x
d
x
{\displaystyle \ \forall A>0\ \exists \int \limits _{0}^{A}{\frac {f(x)}{x}}\,dx}
∃
lim
x
→
+
∞
f
(
x
)
<
+
∞
{\displaystyle \exists \lim \limits _{x\to +\infty }f(x)<+\infty \ }
∫
0
∞
f
(
α
x
)
−
f
(
β
x
)
x
d
x
=
f
(
+
∞
)
ln
(
α
β
)
(
α
>
0
,
β
>
0
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {f(\alpha x)-f(\beta x)}{x}}\,dx=f(+\infty )\ln {\biggl (}{\frac {\alpha }{\beta }}{\biggr )}\ \ (\alpha >0,\beta >0)\ }
Gràcies a la integral de Frullani i amb l'ajut de transformacions elementals, diferenciació i integració respecte a un paràmetre, es poden reduir moltes altres integrals impròpies.
∫
0
∞
sin
(
α
x
)
α
x
−
sin
(
β
x
)
β
x
x
d
x
=
ln
(
β
α
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {{\frac {\sin(\alpha x)}{\alpha x}}-{\frac {\sin(\beta x)}{\beta x}}}{x}}\,dx\,=\,\ln \left({\frac {\beta }{\alpha }}\right)}
∫
0
∞
sin
(
α
x
+
m
)
−
sin
(
β
x
+
m
)
x
d
x
=
sin
(
m
)
⋅
ln
(
β
α
)
{\displaystyle \int \limits _{0}^{\infty }\!{\frac {\sin \left(\alpha x+m\right)-\sin \left(\beta x+m\right)}{x}}{dx}=\sin(m)\cdot \ln \left({\frac {\beta }{\alpha }}\right)}
∫
0
∞
cos
(
α
x
+
m
)
−
cos
(
β
x
+
m
)
x
d
x
=
cos
(
m
)
⋅
ln
(
β
α
)
{\displaystyle \int \limits _{0}^{\infty }\!{\frac {\cos \left(\alpha x+m\right)-\cos \left(\beta x+m\right)}{x}}{dx}=\cos(m)\cdot \ln \left({\frac {\beta }{\alpha }}\right)}
∫
0
∞
m
n
+
α
x
−
m
n
+
β
x
x
d
x
=
m
n
ln
(
β
α
)
{\displaystyle \int \limits _{0}^{\infty }{\frac {{\frac {m}{n+\alpha x}}-{\frac {m}{n+\beta x}}}{x}}\,dx\,=\,{{\frac {m}{n}}\,\ln \left({\frac {\beta }{\alpha }}\right)}}
∫
0
∞
a
r
c
t
g
(
−
α
x
)
α
x
−
a
r
c
t
g
(
−
β
x
)
β
x
x
d
x
=
ln
(
α
β
)
{\displaystyle \int \limits _{0}^{\infty }\!\,{\frac {{\frac {arctg\left(-\alpha \,x\right)}{\alpha \,x}}-{\frac {arctg\left(-\beta \,x\right)}{\beta \,x}}}{x}}{dx}\,={\,\ln \left({\frac {\alpha }{\beta }}\right)}}
Arias de Reyna , Juan «On the Theorem of Frullani » ( Proc. A.M.S. , 109, 1990, pàg. 165-175.Boros , G; Moll , V. Irresistible Integrals (en anglès), 2004, p. 98. Bromwich , T.JAn Introduction to the Theory of Infinite Series (en anglès). Macmillan, 1908, p. 432-433. Frullani , GSopra Gli Integrali Definiti (en italià). XX. Modena: Memorie della Società Italiana delle Scienze, 1828. Ostrowski , A.MProceedings of the National Academy of Sciences , 35(10), octubre 1949, pàg. 612-616. PMC : 1063092 . PMID : 16588938 .Ostrowski , A. MCommentarii Mathematici Helvetici , 51, 1976, pàg. 57–91. DOI : 10.1007/BF02568143 .Tricomi , Francesco GAmerican Mathematical Monthly , 58, 1951, pàg. 158–164. 
![{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,dx&=\int _{0}^{\infty }\left[{\frac {f(xt)}{x}}\right]_{t=b}^{t=a}\,dx\\&=\int _{0}^{\infty }\int _{b}^{a}f'(xt)\,dt\,dx\\&=\int _{b}^{a}\int _{0}^{\infty }f'(xt)\,dx\,dt\\&=\int _{b}^{a}\left[{\frac {f(xt)}{t}}\right]_{x=0}^{x=\infty }\,dt\\&=\int _{b}^{a}{\frac {f(\infty )-f(0)}{t}}\,dt\\&={\Big (}f(\infty )-f(0){\Big )}{\Big (}\ln(a)-\ln(b){\Big )}\\&={\Big (}f(\infty )-f(0){\Big )}\ln {\Big (}{\frac {a}{b}}{\Big )}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd95c908d8d659611307ac944fe8ee04a398fe0b)
![{\displaystyle [b,a]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3015146003c7dab01d939e34e07159fa9604bc3)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle =\left\{\xi \in [\alpha ,\beta ]\Rightarrow \lim \limits _{A\to +0}{A\xi }=0,f(x)\in C[0,+\infty )\Rightarrow \lim \limits _{A\to +0}{f(A\xi )=f(0)}\right\}=f(0)\ln {\biggl (}{\frac {\beta }{\alpha }}{\biggr )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5f19280877a700bea47dbf04e0e8abb49adaa4)
![{\displaystyle =\left\{\rho {\bigl (}\epsilon ,A{\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[\epsilon ,A]\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(x)}{x}}\,dx=F(A)-F(\epsilon )\Rightarrow \int \limits _{\epsilon }^{A}{\frac {f(\alpha x)}{x}}\,dx=F(\alpha A)-F(\alpha \epsilon )\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f443c1915dd05579fda5879f8a988fa0988abb19)
![{\displaystyle =\left\{\rho {\bigl (}A,\Delta {\bigr )}<\infty ,{\frac {f(x)}{x}}\in C[A,\Delta ]\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(x)}{x}}\,dx=F(\Delta )-F(A)\Rightarrow \int \limits _{A}^{\Delta }{\frac {f(\alpha x)}{x}}\,dx=F(\alpha \Delta )-F(\alpha A)\right\}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ad4b1c35ee79d9cd19ec3a043bdec46d423c7d4)
![{\displaystyle =\left\{\eta ,\mu \in [\alpha ,\beta ]\Rightarrow \lim \limits _{\epsilon \to +0}\epsilon \eta =0,\lim \limits _{\Delta \to +\infty }\Delta \mu =+\infty ,f(x)\in C[0,+\infty ]\Rightarrow \lim \limits _{\epsilon \to +0}f(\epsilon \eta )=f(0),\lim \limits _{\Delta \to +\infty }f(\Delta \mu )=f(+\infty )\right\}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8c8fc63db219ac97fdbcba37b6ff0cff3895428)
PDF) (en anglès). Proc. A.M.S., 109, 1990, pàg. 165-175.