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Given the function $$f(x)=\int_0^1 \frac{e^{-x^2(t^2+1)}}{1+t^2} dt $$ I wanna calculate $f'(x)$ as simple as possible.

My attempt:

We know that: $erf(x)=\frac{2}{\sqrt \pi}\int_0^xe^{-t^2}dt$

Put $u=t^2+1$ than: $$f(x)=\int_1^2 \frac{e^{-x^2u}}{ u} du=\int_1^2 \frac{e^{-x^2(\sqrt u)^2}}{ u} du=\int_1^2 \frac{(e^{-\sqrt u^2})^{x^2}}{u} du\overset{*}{=}\int_{x^2}^2 \frac{e^{-u}}{ u} du=-\int_2^{x^2} \frac{e^{-u}}{ u} du$$

Not sure if $(*)$ holds and if so, whether it's the gamma function $-\Gamma(0,x)$ or the exponential integral $E_1(x)$

I just need a little hint because I wanted to express it in terms of the error function, if possible :(

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    The definite integral here will be a function of $x$ instead of a number – 2017-01-21

2 Answers 2

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We are dealing with analytic functions over a compact interval, hence for sure

$$\begin{eqnarray*} \frac{d}{dx}\,f(x)=\frac{d}{dx} \int_{0}^{1} e^{-x^2(1+t^2)}\frac{dt}{1+t^2} &\color{red}{=}& \int_{0}^{1}\frac{d}{dx}e^{-x^2(1+t^2)}\frac{dt}{1+t^2}\\&=&\int_{0}^{1}-2x e^{-x^2(1+t^2)}\,dt\\&=&-2xe^{-x^2}\int_{0}^{1}e^{-x^2 t^2}\,dt\\&=&-2e^{-x^2}\int_{0}^{x}e^{-t^2}\,dt\\&=&-\sqrt{\pi}e^{-x^2}\text{erf}(x) \end{eqnarray*} $$ where $\color{red}{=}$ is differentiation under the integral sign.

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    Just a little question why can we change the bound of integration from $-2xe^{-x^2}\int_0^1e^{-x^2t^2}$ to $-2e^{-x^2}\int_0^xe^{-t^2}$ ? – 2017-01-21
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    @pinkpanther5: just use the substitution $t=\frac{1}{x} u, dt = \frac{1}{x}\,du$. – 2017-01-21
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    Thank you very very much, grazie mille – 2017-01-22
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Use Leibnitz's rule for the differentiation under integral sign which is as follows: if $$F(x)=\int_{a(x)}^{b(x)}f(x,t)dt$$then $$\frac{d}{dx}F(x)=\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt +f(x,b(x)).b'(x)-f(x,a(x)).a'(x)$$This will help you.