$\int_{a}^{+\infty} e^{-x^2} dx$

Question

Question If $a>0$, show that $$\int_a^\infty e^{-t^2} \mathrm{d}t = \frac{e^{-a^2}}{2a}\left(1-\frac{1}{2a^2}+f(a)\right)$$ where $0<f(a)<\displaystyle\frac{3}{4a^4}$

My approach: Applying integration by parts twice, $$\int_a^\infty e^{-t^2} \mathrm{d}t=\frac{1}{2}\int_{a^2}^\infty e^{-p}p^{-1/2} \mathrm{d}p\\=\frac{e^{-a^2}}{2a}\left(1-\frac{1}{2a^2}\right)+\frac{3}{8}\int_{a^2}^\infty e^{-p}p^{-5/8} \mathrm{d}p$$ Observation

$p>a^2\\\implies 0<e^{-p}<e^{-a^2}\\\implies 0<e^{-p}p^{-5/2}<e^{-a^2}a^{-5}$
$p>a^2\\\implies 0<e^{-p}<e^{-a^2}\\\implies 0<e^{-p}p^{-1/2}<e^{-a^2}a^{-5}\\\implies 0<e^{-p}p^{-1/2}p^{-2}<e^{-a^2}a^{-5}p^{-2}\\\implies 0<\int_{a^2}^\infty e^{-p}p^{-5/2} \mathrm{d}p<e^{-a^2}a^{-1}\int_{a^2}^\infty p^{-2} \mathrm{d}p\\\implies 0<\int_{a^2}^\infty e^{-p}p^{-5/2} \mathrm{d}p<e^{-a^2}a^{-3}$

After applying integration by parts twice, I don't know where to go?

Any kinds of help will be appreciated. Thank you!!

@user12986714, a typo. One can easily edit the title after seeing the task. It happens. — Matcha Latte, May 02 '20 at 17:06
@user12986714 Sorry , I copied the integration mathjax from another post. — PNDas, May 02 '20 at 17:07
I can't rotate my computer and I already have a lack of space on it. — Matcha Latte, May 02 '20 at 17:09
If you can't post a right-side up picture, then type your question. Images are discouraged on this site anyway. Images can't be browsed and are not accessible to those using screen readers. — saulspatz, May 02 '20 at 17:14
No, I'm not suggesting that. Just bear it in mind for future posts. Now that the someone has rotated the image, the post is legible. — saulspatz, May 02 '20 at 17:17
I edited the post, I rotated the image. I need an approval now. — Axel, May 02 '20 at 17:20
Here's a related question that may be useful. It doesn't go all the way to solving your problem, but perhaps it will give you some ideas. https://math.stackexchange.com/questions/28751/proof-of-upper-tail-inequality-for-standard-normal-distribution — saulspatz, May 02 '20 at 17:29

score 1 · Accepted Answer · answered May 02 '20 at 17:36

You can continue your solution by applying integration by parts one more time $$\int_{a^2}^{\infty}e^{-x}x^{-5/2}\,dx=\frac{e^{-a^2}}{a^5}-\frac 52\int_{a^2}^{\infty}e^{-x}x^{-7/2}\,dx<\frac{e^{-a^2}}{a^5} $$ Hence, $$0<\frac 38\int_{a^2}^{\infty}e^{-x}x^{-5/2}\,dx<\frac{3e^{-a^2}}{8a^5} $$ and now you can conclude.

$\int_{a}^{+\infty} e^{-x^2} dx$

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