Is there a closed form expression of $\int_0^T \dot f(t) / t dt$?

Question

Let $f(t)$ be a function, $t > 0$

Then is there a closed form expression for,

$$\int_0^T \dfrac{\dot f(t)}{ t} dt$$

My attempt:

$$\int_0^T \dfrac{\dot f(t)}{ t} dt = \int_0^T \dfrac{1}{ t} df(t)$$

Therefore, $$\int_0^T \dfrac{1}{ t} df(t) = \ln(f(T)) - \ln(f(0)) = \ln(f(T)/f(0))$$

Is this correct?

Answer 1

Then is there a closed form expression for, $$\int_0^T \dfrac{\dot f(t)}{t} dt$$

In general not. Take for example $f=-\cos$ which gets you sinus integralis (Si). It is known that it cannot be represented by elementary functions. There are many other examples like for $f=\exp$ or $f(x)=x\ln x -x$ (so that $\dot f(x)=\ln x$).

Answer 2

Your last equality is not correct, for example try $f'(t)=t$. If you perform integration by parts then you get $\displaystyle \int_0^T \frac{f'(t)}{t} \, dt = \left. \frac{f(t)}{t} \right \vert_0^T + \int_0^T \frac{f(t)}{t^2} \, dt$. Or if you further assume that $f''(t)$ exists and integrable you can also get $\displaystyle \int_0^T \frac{f'(t)}{t} \, dt = f''(t) \ln(t) \bigg\vert_0^T - \int_0^T f''(t) \ln(t) \, dt$.

I guess in general there is no "closed form" because if there was a such formula then we could use it express $\displaystyle \int_0^T \frac{\sin t}{t} \, dt$, namely "sinc function" and other integrals.

Answer 3

No, it is not correct. You confuse

$$\int\frac{df}{t}=\,?$$ with

$$\int\frac{df}{f}=\log f+c.$$

As a simple counterexample, take $f(t)=t^2$ and

$$\int\frac{2t}tdt=2t+c\ne\log t^2+c.$$

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There is no way to perform the requested integration without having the particular expression of $f$.