Is there a general way to find an integral representation of a function?

Question

$\underline{\text{Motivation}:-}$

This question really came to my mind when I saw this question where OP is asking for the integral representation of a function he created. This got me thinking, are there general ways to represent functions as integrals?

$\underline{\text{My Question}:-}$

My question is really simple-

Is there a general way to find out the integral representations of a function?

Or do we just keep finding identities involving the function until we just happen to find one involving integrals?

You can clarify using some examples.

I already know about the Abel-plana formula.

Edit:

For example we can write the zeta function as this

$\textstyle\displaystyle{\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+2\int_{0}^{\infty}\frac{\sin(s\tan^{-1}(t)}{(1+t^2)^\frac{s}{2}(e^{2\pi t}-1)}dt}$

Basically what I mean by integral representation of some function $f(x)$ is-

$\textstyle\displaystyle{f(x)=g(x)+\int_{a(x)}^{b(x)}h(x,t)dt}$

Here, $f(x), g(x), a(x), b(x)$ and $h(x,t)$ are all continuous functions.

I don't want obvious answers like-

$g=f$ and $a=b$

Or $g=0$ and $b(x)=x$ and $a(x)=\rho$ such that $f(\rho)=0$ and $h(x,t)=f'(t)$. You can just put these in the equation and realize that this is just FTC.

You can just feel it with your heart that these are not interesting answers. I basically want interesting answers, for example the Abel-plana formula is an interesting one, but since I already know it, there is no reason to repeat.

To be honest I don't know how to clarify more than just saying I want an interesting answer.

Answer 1

There is a versatile formula that I learned from:

How to evaluate $$\sum\limits_{x=0}^\infty \text{erfc}(x)= 1.1619990479471263635323…?$$

where the answer from @StevenClark was good, but the most creative answer was from @John Barber with one solution as the following:

In this sum of integrals, the interval $[0,1)$ will be counted only once, in the $n = 0$ term. The interval $[1,2)$ will be counted twice, in the $n = 0$ and $n = 1$ terms. And so on. This means we can write: $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{2}{\sqrt{\pi}}\int_0^{\infty}\, \lfloor t+1\rfloor \,e^{-t^2}\, dt+1+\frac{2}{\sqrt\pi}\int_1^\infty \lfloor x\rfloor e^{-x^2} dx$$

Therefore a general formula for an integral representation for a function defined by a sum which works for many functions:

$$\sum_{x=0}^\infty f(x)=\int_0^\infty \lfloor x-a\rfloor df(x)=\int_0^\infty \lfloor x-a\rfloor f’(x) dx\implies \int_0^\infty \lfloor x-a\rfloor f(x)dx=\sum_{x=0}^\infty\int f(x)dx\mathop=^{\int f(x)dx=F(x)}\sum_{x=0}^\infty F(x)$$

where every situation that I ever used this formula had $a\in\{-1,0,1\}$

Here is another example using the Modified Bessel Function of the Second Kind and Kelvin functions $\text{kei}(x),\text{ker}(x)$ with this computation

$$\sum_\Bbb N(\text{ker(x)}+ i\text{kei}(x))=\sum_\Bbb N \text K_0\left(\sqrt i x\right) =\sqrt[4]{-1}\int_0^\infty\lfloor x\rfloor \text{kei}_1(x)dx+\sqrt[-4]{-1}\int_1^\infty \lfloor x\rfloor \text{ker}_1(x)dx=(-1)^{-\frac34}\int_1^\infty \lfloor x\rfloor \text K_1\left(\sqrt[4]{-1} x\right)dx= 0.133691752819604391549325780771600891… - 0.725631207729182631737443031218031025… i $$

which has a more traditional integral representation in the question, but that does not use this floor function method. I will add more new examples later. You can also put it in terms of other functions using these floor function identities. This method does not work if the derivative of $f(x)$ has a term canceled. Please correct me and give me feedback!