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I'm a college Sophomore majoring in math and over the summer I've been playing around with improper integrals, specifically integrals with limits at infinity because they've always fascinated me. The highest calculus course I've taken is Calc II, so I might be missing something here.

Anyways, what really got me started on this was my discovery of the Pi function (Gamma -1) accidentally.

I was doing integrals that I made up off the top of my head and I found that

[itex] \int_{0}^{\infty}t^ne^{-t}dt =n! [/itex]

I then realized this integral could actually represent the factorial function for any value of n.

So I asked myself, is there a way to represent ANY real function as an improper integral?

Is there a function [itex]\omega (t,x) [/itex] such that any real function and a fixed lower bound can be represented by [itex]f(x)=\int_{a}^{\infty}\omega (t,x)dt[/itex] ?

I played around with a lot more integrals until I found that [itex]\int_{0}^{\infty}e^{-t/x}dt=x [/itex] for x>0

Since and x>0 will give back any x, any f(x)>0 should give back f(x).

So, this is what I propose can represent any real function f(x) as an improper integral

[itex] f(x)=

\begin{Bmatrix}

\int_{0}^{\infty}e^{-t/f(x)}dt, f(x)>0 & \\

0, f(x)=0 & \\

\int_{0}^{\infty}-e^{t/f(x)}dt, f(x)<0 &

\end{Bmatrix}

[/itex]

The minus sign on the last row is supposed to be outside the integral.

So, an example

[itex] sin(x)=

\begin{Bmatrix}

\int_{0}^{\infty}e^{-t/sin(x)}dt, 2n\pi<x<2n\pi +\pi & \\

0, x=n\pi & \\

\int_{0}^{\infty}-e^{t/sin(x)}dt, 2n\pi-\pi<x<2n\pi &

\end{Bmatrix}

[/itex]

I seem to have found a way to represent any real function as an improper integral as long as I know when it’s greater, equal to, and less than 0.

But my question to you guys is, is this result simply trivial? Can this be used to represent real functions the way I want to? I’m not a very experienced mathematician and I’m just asking your advice.

Anyways, what really got me started on this was my discovery of the Pi function (Gamma -1) accidentally.

I was doing integrals that I made up off the top of my head and I found that

[itex] \int_{0}^{\infty}t^ne^{-t}dt =n! [/itex]

I then realized this integral could actually represent the factorial function for any value of n.

So I asked myself, is there a way to represent ANY real function as an improper integral?

Is there a function [itex]\omega (t,x) [/itex] such that any real function and a fixed lower bound can be represented by [itex]f(x)=\int_{a}^{\infty}\omega (t,x)dt[/itex] ?

I played around with a lot more integrals until I found that [itex]\int_{0}^{\infty}e^{-t/x}dt=x [/itex] for x>0

Since and x>0 will give back any x, any f(x)>0 should give back f(x).

So, this is what I propose can represent any real function f(x) as an improper integral

[itex] f(x)=

\begin{Bmatrix}

\int_{0}^{\infty}e^{-t/f(x)}dt, f(x)>0 & \\

0, f(x)=0 & \\

\int_{0}^{\infty}-e^{t/f(x)}dt, f(x)<0 &

\end{Bmatrix}

[/itex]

The minus sign on the last row is supposed to be outside the integral.

So, an example

[itex] sin(x)=

\begin{Bmatrix}

\int_{0}^{\infty}e^{-t/sin(x)}dt, 2n\pi<x<2n\pi +\pi & \\

0, x=n\pi & \\

\int_{0}^{\infty}-e^{t/sin(x)}dt, 2n\pi-\pi<x<2n\pi &

\end{Bmatrix}

[/itex]

I seem to have found a way to represent any real function as an improper integral as long as I know when it’s greater, equal to, and less than 0.

But my question to you guys is, is this result simply trivial? Can this be used to represent real functions the way I want to? I’m not a very experienced mathematician and I’m just asking your advice.

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