Find $f(x)$ for which $I_{\varphi}(k) = \int_0^{k\varphi} f(x)\,\mathrm{d}x = 1$ As the question says, the problem is as follows:
Question 1 Find a function $f \in \mathcal{C}^{\infty}([0, \infty))$ (or $f \in \mathcal{C}^{\infty}(\mathbb{R})$), such that $$\forall k \in \mathbb{N} : I_{\varphi}(k) =\int_0^{k\varphi} f(x)\,\mathrm{d}x = 1,$$ where $\varphi \in \mathbb{R}_+$ is an arbitrary constant one must choose at the beginning. This means one can choose any constant but then it has to be the same for all $k$.

To rephrase the question. You have to find a continuous, infinitely-many times differentiable function, such that if we choose a ''step'' $\varphi$, no matter how many steps we make, the integral from 0 to our current position is 1.

Example: Function $f(x) = \frac{1}{\pi}\cos^2(x)$ is element of $\mathcal{C}^{\infty}(\mathbb{R})$ and it holds that $$\int_0^{2\pi} f(x) = 1,$$ but $$\int_0^{k\cdot2\pi} f(x) \neq 1,$$ for all $k > 1$. In fact, it holds that $$\int_0^{k\pi} \cos^2(x) \mathrm{d}x = \frac{k\pi}{2},$$ so considering a function $$\tilde{f}(x) = f(k, x) = \frac{2}{k\pi} \cos^2(x)$$ one can clearly see that it would satisfy the problem given, but this function is not OK, since it is a function of $k$ and $x$, not $x$ only.


Question 2 The problem is the same but $f(x)$ can be any function (not necessarily differentiable and continuous) but following condition must hold: $$\mathrm{card}(\Omega) \leq \mathrm{card}(\mathbb{N}),$$ where $\Omega = \{x \in \mathbb{R} : f(x) = 0\}$.

Example: Function $$f(x) = \begin{cases} 1; & x \in [0,1]\\0; & \mathrm{otherwise}\end{cases}$$ is not OK, since $\mathrm{card}(\Omega) = \mathrm{card}(\mathbb{R}) > \mathrm{card}(\mathbb{N})$.


Question 3 Omit all conditions on $f(x)$. In this case, a function from example in Question 2 is OK. In fact, many step functions are now OK.

Question 4 Prove that such function $f$ from Question 1 or Question 2 does not exist.
Have fun with this problem! :)
 A: I assume that the OP meant $\mathcal{C}^\infty\big([0,\infty)\big)$, rather than $\mathcal{C}^\infty\big([0,\infty]\big)$ (the answer would definitely change if I was wrong about this).  I am answering Questions 1, 2, and 4 simultaneously. 
Let $b_\varphi:\mathbb{R}\to\mathbb{R}_{\geq 0}$ be a bump function with support $\left[\frac{\varphi}{2},\varphi\right]$.  Define
$$B_\varphi(x):=\frac{\int_{-\infty}^x\,b_\phi(t)\,\text{d}t}{\int_{-\infty}^{+\infty}\,b_\varphi(t)\,\text{d}t}\text{ for every }x\in\mathbb{R}\,.$$
Consider the function
$$\psi(x):=\big(1-B_\varphi(x)\big)+B_\varphi(x)\,\cos\left(\frac{2\pi}{\varphi}x\right)\text{ for each }x\in\mathbb{R}\,.$$
Then, $f:\mathbb{R}\to\mathbb{R}$ defined by
$$f(x):=\frac{\psi(x)}{\int_0^\varphi\,\psi(t)\,\text{d}t}\text{ for all }x\in\mathbb{R}$$
satisfies the smoothness condition, the integral condition, and the countability condition on $f^{-1}(0)$, as
$$f^{-1}(0)=\varphi\,\mathbb{Z}_{>0}=\Biggl\{\left(k+\frac{1}{2}\right)\varphi\,\Big|\,k\in\mathbb{Z}_{>0}\Biggr\}\,.$$
P.S. It is an interesting challenge to show whether there exists an analytic function $f:\mathbb{R}\to\mathbb{R}$ (or $f:\mathbb{R}_{\geq 0}\to\mathbb{R}$) obeying the integral condition.  I conjecture that the answer is negative.
