Use a function to define an optional parameter

No this is not possible. For a parameter to be optional the value must be a compile time constant. You can however overload the method like so:

private int ChangeC(int a, int b)
{
    return a + b;
}

public void ExampleMethod(int a, int b, int c) {}

public void ExampleMethod(int a, int b)
{
    ExampleMethod(a, b, ChangeC(a, b));
}

This way you don't have to deal with nullable value types

One of the ways:

private int ChangeC(int a, int b)
{
    return a+b; 
} 

public void ExampleMethod(int a, int b, int? c=null)
{
    c = c ?? ChangeC(a,b);
}

Is it possible to use the return value of a function instead of a specific value as optional parameter in a function?

No. It is not possible. The C# programming guide on Optional Arguments says:

A default value must be one of the following types of expressions:

• a constant expression;

• an expression of the form new ValType(), where ValType is a value type, such as an enum or a struct;

• an expression of the form default(ValType), where ValType is a value type.

See other answers for alternative solutions.