Frullani integral Last updated June 20, 2025  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}}.} If f ( ∞ ) {\displaystyle f(\infty )} ∫ c ∞ f ( x ) x d x {\displaystyle \int _{c}^{\infty }{\frac {f(x)}{x}}dx} c > 0 {\displaystyle c>0} ∫ 0 ∞ f ( a x ) − f ( b x ) x d x = − f ( 0 ) ln  a b . {\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\,{\rm {d}}x=-f(0)\ln {\frac {a}{b}}.} 
Proof for continuously differentiable functions 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]} 
Ramanujan, using his master theorem , gave the following generalization. [ 1]   [ 2]  
Let f , g {\displaystyle f,g} [ 0 , ∞ ] {\displaystyle [0,\infty ]} f ( x ) − f ( ∞ ) = ∑ k = 0 ∞ u ( k ) ( − x ) k k !  and  g ( x ) − g ( ∞ ) = ∑ k = 0 ∞ v ( k ) ( − x ) k k ! {\displaystyle f(x)-f(\infty )=\sum _{k=0}^{\infty }{\frac {u(k)(-x)^{k}}{k!}}\quad {\text{ and }}\quad g(x)-g(\infty )=\sum _{k=0}^{\infty }{\frac {v(k)(-x)^{k}}{k!}}} u ( x ) {\textstyle u(x)} v ( x ) {\textstyle v(x)} f {\textstyle f} g {\textstyle g} [ 0 , ∞ ) {\textstyle [0,\infty )} f ( 0 ) = g ( 0 ) {\textstyle f(0)=g(0)} f ( ∞ ) = g ( ∞ ) {\textstyle f(\infty )=g(\infty )} a , b > 0 {\textstyle a,b>0} 
lim n → 0 ∫ 0 ∞ x n − 1 { f ( a x ) − g ( b x ) } d x = { f ( 0 ) − f ( ∞ ) } { log  ( b a ) + d d s ( log  ( v ( s ) u ( s ) ) ) s = 0 } {\displaystyle \lim _{n\rightarrow 0}\int _{0}^{\infty }x^{n-1}\{f(ax)-g(bx)\}dx=\{f(0)-f(\infty )\}\left\{\log \left({\frac {b}{a}}\right)+{\frac {d}{ds}}\left(\log \left({\frac {v(s)}{u(s)}}\right)\right)_{s=0}\right\}} 
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