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I was studying definite integrals and there is a property in my textbook: If $F (t)=\int_a^b g (x,t)dx $ , then $\frac {dF}{dt}= \int_a^b \frac {\partial g (x,t)}{\partial t}$

This makes sense to me as when we integrate, all the $x's $ are going to remain intact. That is, they are independent of $x$ so if we differentiate w.r.t $t $ it will not affect any of the $x$.

But beneath this property there was a question: Evaluate $\int_0^1 \frac {x^b-1}{lnx} dx $ $(b\ge0) $

The solution to it is given as: take the given function as a function of $b $ Then apply the above theorem to simplify it.

Let $f (b)=\int_0^1 \frac {x^b-1}{lnx} dx $

$f'(b)= \int_0^1 \frac {x^b lnx}{lnx} $ Integrating this further gives us the answer.

But differentiating $f (b) $ over here seems counterintuitive to me because we are affecting the $x $ which seems counterintuitive.

Could anyone please explain why this holds good like how can we affect $x $ over here?

user185887
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  • What’s wrong with taking $g(x,t) = \frac{x^t - 1}{\ln x}$ in your first statement? – Theoretical Economist Jul 12 '18 at 17:36
  • Why would be so counterintuitive? If I change variable $ f(t)=\int_{a}^{b}\frac{x^t-1}{\log(x)}dx = \int_{a}^{b}g(x,t)dx $ – Davide Morgante Jul 12 '18 at 17:37
  • @DavideMorgante taking $b $ instead of $t $ is not counterintuitive. But when we differentiate, we are changing the terms involving $x $ that is "our function of $x $ in the integrand does not remain the same." This is counterintuitive. – user185887 Jul 12 '18 at 17:44
  • @user185887 Yes, I get it. But you said that the formula that you gave us it's not counterintuitive. Now, you even said that: "This makes sense to me as when we integrate, all the $t'$ are going to remain intact. That is, they are independent of $x$ so if we differentiate w.r.t $t$ it will not affect any of the $x$" but I say that this is not true and you gave us a clear example! So the differentiation can effect even the $x$ variable – Davide Morgante Jul 12 '18 at 17:48
  • @DavideMorgante could you please provide me a proof ( simple or rigorous) of If $F (t)=\int_a^b g (x,t)dx $ , then $\frac {dF}{dt}= \int_a^b \frac {\partial g (x,t)}{\partial t}$ – user185887 Jul 12 '18 at 17:53

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The property that you mention is a simpler case of Leibnitz's integral rule which states that $$\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f(x,t)dx\right) = f(b(t),t)\frac{d}{dt}b(t)-f(a(t),t)\frac{d}{dt}a(t) +\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x,t)dx$$

We want to sketch a proof for your simpler version, where $a(t)$ and $b(t)$ are constant. Let's begin.

First we define $$g(t) = \int_a^bf(x,t)dx$$ the total derivative with respect to $t$ of this function by definition is $$g'(t)=\lim_{h\rightarrow\ 0}\frac{g(t+h)-g(t)}{h} = \lim_{h\rightarrow\ 0}\frac{\int_a^bf(x,t+h)dx-\int_a^bf(x,t)dx}{h}$$ for the linearity of the integral $$\lim_{h\rightarrow\ 0}\int_a^b\frac{f(x,t+h)-f(x,t)}{h}dx$$ if the function $f(x,t)$ is well behaved we can pass the limit in the integral and get $$\int_a^b\lim_{h\rightarrow\ 0}\frac{f(x,t+h)-f(x,t)}{h}dx = \int_a^b\frac{\partial}{\partial t}f(x,t)dx = \frac{d}{dt}g(t) = \frac{d}{dt}\int_a^b f(x,t)dx$$

Davide Morgante
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