of x in [a, b], On the other hand, Hence φ(α) is a continuous function. Similarly if exists and is continuous, then for all ε > 0 there exists Δα such that: Therefore, where Now, ε → 0 as Δα → 0, so This is the formula we set out to prove. Now, suppose where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then, A form of the mean value theorem, where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in Dividing by Δα, letting Δα → 0, noticing ξ1 → a and ξ2 → b and using the above derivation for yields This is the general form of the Leibniz integral rule. Examples Example 1: Fixed limits Consider the function The function under the integral sign is not continuous at the point (x, α) = (0, 0), and the function φ(α) has a discontinuity at α = 0 because φ(α) approaches ±π/2 as α → 0±. If we differentiate φ(α) with respect to α under the integral sign, we get which is, of course, true for all values of α except α = 0. This may be integrated (with respect to α) to find Example 2: Variable limits An example with variable limits: Applications Evaluating definite integrals The formula can be of use when evaluating certain definite integrals. When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration. Example 3 Consider Now, As varies from to , we have Hence, Therefore, Integrating both sides with respect to , we get: follows from evaluating : To determine in the same manner, we should need to substitute in a value of greater than 1 in . This is somewhat inconvenient. Instead, we substitute , where . Then, Therefore, The definition of is now complete: The foregoing discussion, of course, does not apply when , since the conditions for differentiability are not met. Example 4 First we calculate: The limits of integration being independent of , we have: On the other hand: Equating these two relations then yields In a similar fashion, pursuing yields Adding the two results then produces which computes as desired. This derivation may be generalized. Note that if we define