Differentiation under the Integral Sign

Differentiation and integration are two central operations in calculus. There are many results on differentiation and integration treating them individually. But only few results tell about the relation between them . One of them is the fundamental theorem of calculus which says that differential of the integral of a continuous gives back the same function. But what about functions of two variables?


Suppose f(x,y) is a function of two variables, then when can we say

     `d /dx` `int_a^bf(x,y)dy` =  `int_a^b` `del/(delx)` `f(x,y)``dy`       ?

Above question is well answer by " Leibniz's rule for differentiation under integral sign ". It is named after Gottfried Leibniz, a well reputed mathematician.


Leibniz's Rule for Differentiation under Integral Sign :


Statement :  Let  f : [a , b] `xx` [c , d] `->` `RR` be a function such that `f` and   `(delf)/(delx)`are continous on the rectangle [a , b] `xx` [c , d]  then,

`d/(dx)``int_c^df(x,y)dy` = `int_c^d` `del/(delx)``f(x,y) dy`

Proof :  Let  h(x) = `int_c^df(x,y)dy` .  So we need to show that  hl(x) = `int_c^d` `del/(delx)``f(x,y) dy`

By definition, hl(x) = `lim_(t->0)` `( h(x+t) - h(x))/t`   .

Now consider the numerator, h(x + t) - h(x) = `int_c^df(x+t,y)dy` -  `int_c^df(x,y)dy`

                                                                                                  = `int_c^d(f(x+t,y) - f(x,y))dy`


So  hl(x) = `lim_(t->0)``int_c^d` `(f(x+t,y) - f(x,y))/t dy` .  Since  f is continous on [a , b] `xx` [c , d],  we can apply mean value theorem.    So there exist t0`in` (0 , t) with t0 `->` 0 as t `->` 0, such that  f(x + t, y) - f(x , y) = `del/(delx)`f(x+t0 , y) .

                      So we have hl(x) = `lim_(t->0)` `int_c^d``del/(delx)``f(x+`t0`,y) dy` . Also since `del/(delx) f(x,y)`  is continous on [a ,b] `xx` [c ,d], which is a compact subset  of  `RR`2 ,  `del/(delx) f(x,y)` is uniformly continous and hence we can interchange  the `lim_(t->0)` and `int_c^d` .

                      So we get  hl(x) = `int_c^d``lim_(t->0)``del/(delx)``f(x+`t0`,y) dy` . But as t `->` 0, t0 `->` 0, and since `(delf)/(delx) ` is continous,

                                     we get hl(x) = `d/dx` `int_c^df(x,y)dy` =  `int_c^d``del/(delx)f(x,y) dy` .  This completes the proof the rule.


An Example Showing Differentiation under Integral Sign

  • Find `d/dx` `int_0^pi` `ln ( 1 - 2x cos(y)` `+` `x`2 `)``dy`

      Solution :

      Directly apply the above rule.  We get `int_0^pi` `del/(delx)``(ln(1 - 2x cos(y) + x`2`) ) dy`

`int_0^pi` `(2x - 2cos(y) )``/` `( 1 - 2xcos(y) + x`2`) dy`

 On simplification we get it as `(2pi)/x` .


A more Generalised Rule for Differentiation under Integral Sign



A more general form of the above rule for differentiation is also available. Here the boundaries are also functions of x, that is we have c = c(y)  and d = d(y). Then the rule is

`d/dy` `int_(c(y))^(d(y))f(x,y)dx`  = `(dd(y))/dy` .`f(d(y),y)` `-` `(dc(y))/dy``f(c(y),y)` `+` `int_(c(y))^(d(y))``del/(dely)``f(x,y)`  `dx`