Define

We establish two fixed-point theorems for mappings satisfying a 
general contractive inequality of integral type. These results substantially extend the theorem of Branciari (2002).

In a recent paper [1], Branciari established the following theorem.Theorem 1.Let (X, d) be a complete metric space, c ∈ [0, 1), f : X → X a mapping such that, for each x, y ∈ X, d(f x,f y) where ϕ : R + → R + is a Lebesgue-integrable mapping which is summable, nonnegative, and such that, for each > 0, 0 ϕ(t)dt > 0. Then f has a unique fixed point z ∈ X such that, for each x ∈ X, lim n f n x = z.
In [1], it was mentioned that (1) could be extended to more general contractive conditions.It is the purpose of this paper to make such an extension to two of the most general contractive conditions.Define Our first result is the following theorem.
Theorem 2. Let (X, d) be a complete metric space, k ∈ [0, 1), f : X → X a mapping such that, for each x, y ∈ X, where ϕ : R + → R + is a Lebesgue-integrable mapping which is summable, nonnegative, and such that Then f has a unique fixed point z ∈ X and, for each x ∈ X, lim n f n x = z.
One would like to be able to replace (2) with the integral form of Ćirić's condition [3], that is, where

M(x, y) := max d(x,y),d(x,f x),d(y,f y),d(x,f y),d(y,f x) . (26)
But this is not possible since, as the following example shows, one must assume that the orbits are bounded.
Then, for n > m, where t := n − m.Note that, for any t ∈ N, Since ϕ(t) = φ (t), it follows from (28) that or, equivalently, and ( 25) is satisfied.However, the orbits are not bounded and f has no fixed points.
Theorem 1 is clearly a special case of Theorem 2. With ϕ equal to the constant function 1, Theorem 2 reduces to [2,Theorem 2.5] It is possible to prove a weaker theorem involving condition (25).
For any set A, δ(A) will denote the diameter of A. Theorem 4. Let (X, d) be a complete metric space, k ∈ [0, 1), f : X → X a mapping such that, for each x, y ∈ X, (25) is satisfied, where ϕ : R + → R + is a Lebesgue-integrable mapping which is summable, nonnegative, and satisfies (4).If there exists a point x ∈ X with bounded orbit, then f has a unique fixed point z ∈ X.
Proof.From the definition of O(x, n), there exist integers i, j satisfying 0 Proof of Claim 5. We may assume that δ(O(x, n)) > 0 for each n, since, if there exists an n for which δ(O(x, n)) = 0, then f has a fixed point.
Pick an x ∈ X with bounded orbit.Let m and n be integers with m > n.Then, from (25), which implies that z = w, and the fixed point is unique.
The following example shows that ( 2) is indeed a proper extension of (1).