Problem of the Day 3

Relevant wiki: Error function

$\begin{aligned} I & = \int_0^1 \left(1+e^{-x^2}\right) dx \\ & = \int_0^1 dx + \color{#3D99F6} \int_0^1 e^{-x^2} dx & \small \color{#3D99F6} \text{Note that error function }\text{erf } (x) = \frac 2{\sqrt \pi} \int_0^x e^{-t^2} dt \\ & = 1 + \frac {\sqrt \pi}2 \text{erf }(1) \\ & \approx 1.74682 \end{aligned}$

Therefore, the answer is " None of the others ".

For the interval $x \in (0,1)$ , we can see that $e^x > e^{x^2} \implies e^{-x} < e^{-x^2}$ and since $f(t) = e^t$ is always positive and an increasing function, therefore $\displaystyle \int_0^1 (1+e^{-x^2}) \,dx > \int_0^1 (1+e^{-x}) \,dx \implies \displaystyle \int_0^1 (1+e^{-x^2}) \,dx > 2 - e^{-1}$ . Also note that $e^{-x^2} < 1 \ \forall \ x \in \mathbb{R}^{+}$ , thus $\displaystyle \int_0^1 (1+e^{-x^2}) \,dx < \int_0^1 (1+1) \,dx \implies \displaystyle \int_0^1 (1+e^{-x^2}) \,dx < 2$ . Hence $2 - e^{-1} < I < 2$ and none of the options given even lie inside the interval, giving us the answer $\boxed{\text{None of these}}$ .

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The integral $\displaystyle \int e^{-x^2} \,dx$ is found to have no antiderivative in terms of elementary functions. One can simply use the Maclaurin series for $e^t$ to get the result as $\displaystyle \int_0^z e^{-x^2} \,dx = \int_0^z \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{n!} \,dx = \sum_{n=0}^{\infty} \int_0^z \dfrac{(-1)^n x^{2n}}{n!} \,dx = \sum_{n=0}^{\infty} \dfrac{(-1)^n z^{2n+1}}{(2n+1) n!}$ which gives quick convergence for $z=1$ and the answer can thus be approximated.

see error function

$\displaystyle\int_0^1 (1+e^{-x^2})\text{dx} = 1+\displaystyle\int_0^1 e^{-x^2} \text{dx}$

$= 1+ \displaystyle\int_0^1 \frac{1}{2\sqrt{z}}e^{-z} \text{dz}$ Substituting $z=x^2$

$= 1+ \frac{1}{2} \displaystyle\int_0^1 z^{-\frac{1}{2}} e^{-z} \text{dz}$

$= 1+ \frac{\sqrt{\pi}}{2} \displaystyle\int_0^1 \frac{z^{1-\frac{1}{2}}}{\Gamma(\frac{1}{2})} e^{-z} \text{dz}$

$= 1+ \frac{\sqrt{\pi}}{2} \text{I}\Big(\frac{1}{2},1\Big)$ where $\text{I}\Big(\frac{1}{2},1\Big)$ denotes $\text{The Incomplete Gamma Function}$

$\text{So the answer is "None of the others"}$

We can approximate using taylor expansion..........