If $\displaystyle{g(x)=\int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1}\,dt}$, find $g'(x)$.
Any hint?
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9$\begingroup$ According to Leibniz theorem of differentiation under integral sign, you can push that derivative wrt $x$ inside the integral. $$\frac{d}{dx}\int_{0}^{1}\frac{e^{-x(t^2+1)}}{t^2+1}dt=\int_{0}^{1}\frac{1}{t^2+1}\frac{d}{dx}e^{-x(t^2+1)}dt$$ $\endgroup$– user985108Commented Nov 7, 2021 at 18:29
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$\begingroup$ Thaank you very much :D $\endgroup$– Daniel ChecaCommented Nov 7, 2021 at 18:39
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$\begingroup$ See also the fourth proof here. $\endgroup$– J.G.Commented Nov 13, 2021 at 12:19
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1 Answer
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Let assume: $$\int f(x, t)dt = F(x, t) + c_0$$
$$g(x)=\int_{\phi(x)}^{\psi(x)}{f(x, t)}dt =\left|F(x, t)\right|_{t = \phi(x)}^{t = \psi(x)} = F(x, \psi(x)) - F(x, \phi(x))$$
$$\begin{align*}g'(x)&= \left[\frac {\partial}{\partial x}F(x, \psi(x)) + \frac {\partial}{\partial \psi(x)}F(x, \psi(x)).\color{red}{\psi'(x)}\right]- \left[\frac {\partial}{\partial x}F(x, \phi(x)) + \frac {\partial}{\partial \phi(x)}F(x, \phi(x)).\color{red}{\phi'(x)}\right] \end{align*}$$
$$ g'(x)=\left[\frac {\partial}{\partial x}F(x, \psi(x)) -\frac {\partial}{\partial x}F(x, \phi(x))\right]$$
$$\begin{align*}\color{green}g'(x) = \int_{\phi}^{\psi} \frac {\partial}{\partial x}f(x, t)dt {g'(x)} & = \int_0^1 {\frac{\partial}{\partial x}\frac{e^{-x^2(t^2+1)}}{t^2+1}}dt\\ & = -2x\int_0^1 {\frac{e^{-x^2(t^2+1)}(t^2+1)}{t^2+1}}dt\\ & = -2x\int_0^1e^{-x^2t^2}.e^{{-x^2}}dt\\ & = -2xe^{{-x^2}}\int_0^1e^{-x^2t^2}dt\\ & = -2xe^{-x^2}\times \frac {\sqrt\pi}{2} \times \frac{\text{erf}(x)}{x}\\ & = \color{green}{-\sqrt\pi e^{-x^2} \text{erf}(x) = -e^{-x^2}\gamma(\frac 1{2}, x^2)} \end{align*}$$
