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Type of improper integral with general solution
In mathematics , Frullani integrals are a specific type of improper integral named after the Italian mathematician Giuliano Frullani . The integrals are of the form
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
{\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x}
where
f
{\displaystyle f}
function defined for all non-negative real numbers that has a limit at
∞
{\displaystyle \infty }
f
(
∞
)
{\displaystyle f(\infty )}
The following formula for their general solution holds if
f
{\displaystyle f}
(
0
,
∞
)
{\displaystyle (0,\infty )}
∞
{\displaystyle \infty }
a
,
b
>
0
{\displaystyle a,b>0}
∫
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}}.}
Proof for continuously differentiable functions [ edit ] A simple proof of the formula (under stronger assumptions than those stated above, namely
f
∈
C
1
(
0
,
∞
)
{\displaystyle f\in {\mathcal {C}}^{1}(0,\infty )}
Fundamental theorem of calculus to express the integrand as an integral of
f
′
(
x
t
)
=
∂
∂
t
(
f
(
x
t
)
x
)
{\displaystyle f'(xt)={\frac {\partial }{\partial t}}\left({\frac {f(xt)}{x}}\right)}
f
(
a
x
)
−
f
(
b
x
)
x
=
[
f
(
x
t
)
x
]
t
=
b
t
=
a
=
∫
b
a
f
′
(
x
t
)
d
t
{\displaystyle {\begin{aligned}{\frac {f(ax)-f(bx)}{x}}&=\left[{\frac {f(xt)}{x}}\right]_{t=b}^{t=a}\,\\&=\int _{b}^{a}f'(xt)\,dt\\\end{aligned}}}
and then use Tonelli’s theorem to interchange the two integrals:
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
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 }\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\to \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}}}
Note that the integral in the second line above has been taken over the interval
[
b
,
a
]
{\displaystyle [b,a]}
[
a
,
b
]
{\displaystyle [a,b]}
The formula can be used to derive an integral representation for the natural logarithm
ln
(
x
)
{\displaystyle \ln(x)}
f
(
x
)
=
e
−
x
{\displaystyle f(x)=e^{-x}}
a
=
1
{\displaystyle a=1}
∫
0
∞
e
−
x
−
e
−
b
x
x
d
x
=
(
lim
n
→
∞
1
e
n
−
e
0
)
ln
(
1
b
)
=
ln
(
b
)
{\displaystyle {\int _{0}^{\infty }{\frac {e^{-x}-e^{-bx}}{x}}\,{\rm {d}}x={\Big (}\lim _{n\to \infty }{\frac {1}{e^{n}}}-e^{0}{\Big )}\ln {\Big (}{\frac {1}{b}}}{\Big )}=\ln(b)}
The formula can also be generalized in several different ways.[ 1]
