How to reasonably (numerically) estimate $\int_0^1 (1 - x) \sqrt{2\over\pi} e^{x^2/2}dx$? As the question suggests, I came across the following integral in a calculation:
$$\int_0^1 (1 - x) \sqrt{2\over\pi} e^{x^2/2}dx.$$ According to Wolfram Alpha, this equals$$\text{erfi}\left({1\over{\sqrt{2}}}\right) - (\sqrt{e} - 1)\sqrt{2\over\pi} \approx 0.435834.$$However, I'm wondering if anyone can give a reasonable numerical estimate for this integral from first principles (pencil and paper) without using a calculator or Wolfram Alpha.
I've tried but made little to no progress, and two PhD students I consulted didn't know either, so I'm asking here.
 A: Write
$$(1-x)\, e^{\frac {x^2}2}=\sum_{n=0}^\infty \frac {x^{2n}}{2^n\,n!}-\sum_{n=0}^\infty \frac {x^{2n+1}}{2^n\,n!}$$
$$\int_0^1 (1-x)\, e^{\frac {x^2}2}\,dx=\sum_{n=0}^\infty \frac 1{2^{n+1}\,(2n+1)\, (n+1)!}$$ COmputing the partial sums
$$S_p=\sum_{n=0}^p \frac 1{2^{n+1}\,(2n+1)\, (n+1)!}$$ this generates the sequence
$$\left\{\frac{1}{2},\frac{13}{24},\frac{131}{240},\frac{2447}{4480},\frac{26429}{48384 },\frac{5814401}{10644480},\frac{151174459}{276756480},\cdots\right\}$$
A: $$\int_0^1 (1 - x) \sqrt{2\over\pi} e^{x^2/2}dx=\int_0^1 \sqrt{2\over\pi} e^{x^2/2}dx-\int_0^1 x \sqrt{2\over\pi} e^{x^2/2}dx$$
The first term is imaginary error function. The second term is easier.
Since, $erfi(x)=\frac{2}{\sqrt{\pi}}\int_0^x  e^{t^2}dt$
First term: $erfi(\frac{1}{\sqrt{2}})=\int_0^1 \sqrt{2\over\pi} e^{x^2/2}dx$
Second term: $\int_0^1 x \sqrt{2\over\pi} e^{x^2/2}dx=\int_0^1 \sqrt{2\over\pi} e^{x^2/2}d\frac{x^2}{2}=[\sqrt{2\over\pi} e^{x^2/2}]^1_0=\sqrt{2\over\pi}(\sqrt{e}-1)$
Then the problem is how to estimate $erfi(\frac{1}{\sqrt{2}})$ use calculator.
By trapezoidal rule
$erfi(\frac{1}{\sqrt{2}})\approx \sqrt{2\over\pi}(\frac{1+\sqrt{e}}{2}) \approx 1.0567$
By mid-point rule
$erfi(\frac{1}{\sqrt{2}})\approx \sqrt{2\over\pi}(e^\frac{1}{8}) \approx 0.9041$
Use Maclaurin series
$erfi(\frac{1}{z})\approx \frac{2}{\sqrt{\pi}}(z+\frac{z^3}{3}+\frac{z^5}{10}+\frac{z^7}{42}+\frac{z^9}{216}+...)$
$erfi(\frac{1}{\sqrt{2}})\approx \sqrt{2\over\pi}(1+\frac{1}{6}+\frac{1}{40}+\frac{1}{336}+\frac{1}{3456}+...) \approx 0.9534$
This series can be as accurate as you want and workable by a calculator.
A: $e^{\frac{x^2}{2}} \ge 1 + \frac{x^2}{2}$ gives us a lower bound  of
$$\sqrt{\frac{2}{\pi}} \int_0^1 (1 - x) \left( 1 + \frac{x^2}{2} \right) \, dx = \sqrt{\frac{2}{\pi}} \left( \frac{13}{24} \right) > 0.43.$$
To get a similar upper bound, by convexity we have $e^x \le 1 + (e - 1) x$ for $x \in [0, 1]$ which gives $e^{\frac{x^2}{2}} \le 1 + (e - 1) \frac{x^2}{2}$, hence an upper bound of
$$\sqrt{\frac{2}{\pi}} \int_0^1 (1 - x) \left( 1 + (e - 1) \frac{x^2}{2} \right) \, dx = \sqrt{ \frac{2}{\pi} } \left( \frac{11 + e}{24} \right) < 0.46.$$
If you want more accuracy than this or a way to bound $\sqrt{\frac{2}{\pi}}$ we could keep going but I assume dealing with the integral was the part you wanted to focus on. We can use the next term in the Taylor series expansion which gives bounds $1 + x + \frac{x^2}{2} \le e^x \le 1 + x + (e - 2) x^2$. Using this term the graphs of the functions are already nearly visually indistinguishable.
