I want to apply the Leibniz integration rule twice to this integral:

$$ \frac{\textrm d^2}{\textrm d^2 x}\int_a^x g(s)(x-s)ds $$

The first application gives:

$$ \frac{\textrm d}{\textrm d x}\int_a^x g(s)(x-s)ds = \int_a^x\frac{\partial}{\partial x}\left(g(s)(x-s)\right)ds = \int_a^x g(s)ds $$

But the next one troubles me:

$$ \frac{\textrm d}{\textrm d x}\int_a^x g(s)ds = \int_a^x\frac{\partial}{\partial x}g(s)ds = $$

The result should be

$$ g(x) $$

But I cannot see why $g(x)$ is the result.

Any ideas?


What you wrote is not correct: $$ \frac{d}{dx}\int_a^x g(s) \mathrm ds\ne\int_a^x \frac{d}{dx}g(s)\mathrm ds $$ The result follows from the regular fundamental theorem of calculus $$ \frac{d}{dx}\int_a^xg(s)\mathrm ds=g(x) $$ You may have gotten lucky in the first term and not been clear on how this rule works. In general, $$ \frac{d}{dx}\int_{a(x)}^{b(x)}f(t,x)\mathrm dt=f(b(x),x)b'(x)-f(a(x),x)a'(x)+\int_{a(x)}^{b(x)}f_x(t,x)\mathrm dt $$ So in the first differentiation, you have $$ \frac{d}{dx}\int_a^x g(s)(x-s)\mathrm ds=g(x)(x-x)+\int_a^x g(s)\mathrm ds= \int_a^x g(s)\mathrm ds $$


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