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Consider the step function: $$ S(x)=\begin{cases} m_1 & \ \ x<a \\ m_2 & \ \ x\ge a \end{cases} $$

I've recently come across an integral of the form $$\int_x^\infty \frac{S'(t)}{S(t)}f(t)dt$$ for arbitrary function $f$

My first thought was that the derivative $S'$ will act as a delta function spike at $t=a$ making the integral just be $f(a)/S(a)$ for $x<a$ and $0$ otherwise. However, this isn't the case as the value of $S(a)$ is up to my chosen definition. Since this problem arose from physics where a definite answer is needed, I did some more digging.

Upon making continuous approximations for $S(x)$ as smooth sigmoid functions with a parameter that allows them to asymptotically approach the sharp step function above and numerically evaluating the above, I did not get anything like $f(a)/S(a)$, even when $S(a)=\frac{m_1+m_2}{2}$. With some more numerical calculations, I've found that $$ \int_x^\infty \frac{S'(t)}{S(t)}f(t)dt=f(a)\int_x^\infty \frac{S'(t)}{S(t)}dt=f(a)\ln(S(t))|_x^\infty $$ Finally, this gives for $x<a$: $$ \int_x^\infty \frac{S'(t)}{S(t)}f(t)dt=f(a)\ln\frac{m_2}{m_1} $$ and $0$ otherwise. I am getting this regardless of the choice of $f$ or the sigmoid function used (I have used both hyperbolic tangent and the error function). I am just struggling to prove this analytically. Any help is very much appreciated!

STL
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    $S'/S$ is the derivative of $T := \log(S),$ and so acts like $\log(m_2/m_1) \delta(x-a)$ (since $T$ jumps by $\log(m_2/m_1)$ at $a$). You're computing $\int_x^\infty f(t) T'(t) \mathrm{d}t,$ which then works out to $f(a) \log(m_2/m_1)$ if $x < a$. As for proof - if you're comfortable with the statement that $\int_x^\infty f(t) S'(t) = (m_2 - m_1)f(a)$ for $x < a,$ then this is the same argument. If not, then you should tell us the sense in which you're working (distributions? Gen functions? I ask because you haven't added those tags). – stochasticboy321 Jan 06 '25 at 19:33
  • Are $m_1$ and $m_2$ strictly positive? – md2perpe Jan 06 '25 at 19:58
  • @md2perpe yes, both are strictly positive in addition to $a>0$ as well – STL Jan 06 '25 at 20:31
  • @stochasticboy321 I think this makes sense, thank you! – STL Jan 06 '25 at 20:31

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The OP's analysis of the integral is fine. Since $S(t)$ is a step function, its derivative $S'(t)$ is indeed a delta function. Hence the the integral is only non-zero for values $t$ very close to $t = a$. Therefore it is permitted to set $f(t) = f(a)$, and take this term out of the integral. We thus get:

$$I = \int_x^\infty \frac {S'(t)}{S(t)}f(t)dt = f(a)\int_x^\infty \frac {S'(t)}{S(t)}dt$$

Since the delta function is located at $x=a$, the integral is zero when $a$ is outside of the integration range. Assuming that $x < a$, we can extend the integration to minus infinity. We can now exploit the property that $S'(t)$ is the derivative of $S(t)$:

$$I = f(a) \int_{-\infty}^{\infty}d(log(S(t))) = f(a)[log(S(\infty)-log(S(-\infty))] = f(a)log(m_2/m_1)$$

This result is identical to that obtained by the OP.

M. Wind
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