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Above picture, I want to know this derivation that

$$ \frac{d}{dT}*∫^T_tf(t,u)du = ∫^T_t\frac{∂}{∂T}f(t,u)du + f(t,T)\frac{d}{dT}*T - 0 $$

I think $\frac{d}{dT}$ means total derivative. So to come up with $\frac{∂}{∂T}$ : partial derivative notation, I think, it should also come up with $\frac{∂}{∂t}$! But As you can see it doesn't deal with that term.

What makes me even uncomfortable is that this part : $∫^T_t\frac{∂}{∂T}f(t,u)du + f(t,T)\frac{d}{dT}*T$

$\frac{∂}{∂T}, \frac{d}{dT}$ are alive together! From the only one equation! I mean, it should be something like this below,

$$ \frac{d}{dT}*∫^T_tf(t,u)du = \frac{∂}{∂T} \left(∫^T_tf(t,u)du\right)dT + \frac{∂}{∂t} \left(∫^T_tf(t,u)du\right)dt $$

Additionaly, how the partial derivative get in the integral notation? I mean, $∫^T_t\frac{∂}{∂T}f(t,u)du$. not $\frac{∂}{∂T}∫^T_tf(t,u)du$

gt6989b
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user13232877
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1 Answers1

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Assuming that $t$ and $u$ are independent of $T$, this result follows directly from the fundamental theorem of calculus, part 1 $$ {\mathrm{d} \over \mathrm{d}x}\int_a^x f(t)\mathrm{d}t = f(x). $$

What may be causing your confusion is the correct, but completely unnecessary use of Leibniz integral rule $$ {\mathrm{d} \over \mathrm{d}x}\int_{a(x)}^{b(x)}f(x,t)\mathrm{d}t = f(x,b(x)){\mathrm{d} \over \mathrm{d}x}b(x) - f(x,a(x)){\mathrm{d} \over \mathrm{d}x}a(x) + \int_{a(x)}^{b(x)}{\partial \over \partial x}f(x,t)\mathrm{d}t $$ since in your case $f$ is not a function of $T$.

Victor Souza
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