Is the 'Integer Function' Bound Valid for Given Functions f(t) and g(u)?

  • Thread starter zetafunction
  • Start date
  • Tags
    Integral
In summary, for a given function f(t) and g(u), where u=x/t and g is non-decreasing and positive for positive arguments, the 'integer function' can be expressed as \int_{0}^{\infty}dt [g(x/t)]f(t) and is valid when compared to \int_{0}^{\infty}dt g(x/t)f(t). If f(t) is greater than or equal to 0 for t >= 0, using (g(x/t) + 1) would be a better option. Additionally, if f(t) is expressed as f^{+}(t) - f^{-}(t), where f^{+}(t) and f^{-}(t)
  • #1
zetafunction
391
0
given [x] the 'integer function' would be the following bound valid ??

[tex] \int_{0}^{\infty}dt [g(x/t)]f(t) \le \int_{0}^{\infty}dt g(x/t)f(t) [/tex]

for a given functions f(t) and g(u) u=x/t , here g is a non-decreasing positive function for positive arguments (and real) of parameter u=x/t
 
Physics news on Phys.org
  • #2
Hi,

if f(t) >= 0 for t >=0, then using (g(x/t) + 1) would be a better idea ?
Now if
[tex] f(t) = f^{+}(t) - f^{-}(t) [/tex],
where [tex]f^{+}(t) and f^{-}(t)[/tex] denotes max(f(t),0) and max(-f(t),0) respectively
then one can use g(x/t) + 1 for f+(t) and g(x/t) - 1 for f-(t) to obtain a bound.
 

1. How do you solve an integral?

Solving an integral involves finding the antiderivative of a function and evaluating it at the given limits of integration. This can be done using various methods, such as u-substitution, integration by parts, or trigonometric substitution.

2. What is the purpose of an integral?

An integral is used to find the area under a curve or the accumulation of a quantity over a specific interval. It is also used to solve problems in physics, engineering, and other fields that involve continuous change.

3. Can all functions be integrated?

No, not all functions can be integrated. Some functions, such as non-elementary functions, cannot be integrated using traditional methods. In these cases, numerical methods or approximation techniques may be used.

4. How is an integral different from a derivative?

An integral is the inverse operation of a derivative. While a derivative gives the rate of change of a function at a specific point, an integral gives the total change or accumulation of the function over a given interval.

5. What are some real-world applications of integrals?

Integrals have many real-world applications, such as calculating the area under a curve in physics and engineering problems, finding the volume of irregular shapes in geometry, and determining the average value of a function in statistics. They are also used in economics, biology, and other fields to model and analyze continuous processes.

Similar threads

Replies
4
Views
738
Replies
1
Views
926
  • Calculus
Replies
6
Views
1K
Replies
3
Views
1K
Replies
3
Views
1K
  • Calculus
Replies
3
Views
853
Replies
11
Views
2K
Replies
4
Views
2K
Replies
1
Views
1K
Back
Top