How to show that $\int_{t}^{\infty} e^{-\frac{x^2}{2}} dx \leq \frac{1}{t}e^{-\frac{t^2}{2}}$ where $1 \leq t$?

Question

How to show that

$$ \int_{t}^{\infty} e^{-\frac{x^2}{2}} dx \leq \frac{1}{t}e^{-\frac{t^2}{2}} $$ where $t \geq 1$.

Answer 1

$$t\int_{t}^{\infty} e^{-\frac{x^2}{2}} dx =\int_{t}^{\infty}t e^{-\frac{x^2}{2}} dx \leq \int_{t}^{\infty} xe^{-\frac{x^2}{2}} dx =- e^{-\frac{x^2}{2}}\Big|_{t}^{\infty} =e^{-\frac{t^2}{2}}$$ where $x\geq t \geq 1$.

Answer 2

Hint: Inside the integrand, multiply and divide by $x$. Then as $x>t$, we have $1/x < 1/t$. Hope this helps. Interestingly this holds for any $ t>0$.

Answer 3

This is taken from my answer here: But how to use this to show that $ \ \int_5^{\infty} e^{-x^2} dx \ $ is negligible as compared to $ \int_5^{\infty} e^{-5x} dx \ $?

$(\dfrac1{x}e^{-x^2/2})' =\dfrac1{x}(e^{-x^2/2})'-\dfrac1{x^2}e^{-x^2/2} =-e^{-x^2}-\dfrac1{x^2}e^{-x^2} $ so $\int e^{-x^2/2}dx =-\dfrac1{x}e^{-x^2/2}-\int \dfrac1{x^2}e^{-x^2/2}dx $ so

$\begin{array}\\ \int_a^{\infty} e^{-x^2/2}dx &=-\dfrac1{x}e^{-x^2/2}|_a^{\infty}-\int_a^{\infty} \dfrac1{x^2}e^{-x^2/2}dx\\ &=\dfrac{e^{-a^2/2}}{a}-\int_a^{\infty} \dfrac1{x^2}e^{-x^2}dx\\ &<\dfrac{e^{-a^2/2}}{a}\\ \end{array} $

You can also get a lower bound as shown in that answer.