For \(x>0\), let \[ f(x)=e^{x^2/2}\int_x^\infty e^{-t^2/2}\,dt. \]
- Prove that \[ 0<f(x)<\frac1x. \]
- Prove that \(f\) is strictly decreasing on \((0,\infty)\).
Solution
1. For every \(x>0\),\[ f(x)>0. \]
The integrand \(e^{-t^2/2}\) is positive for every real \(t\), so \[ \int_x^\infty e^{-t^2/2}\,dt>0. \] Multiplication by the positive factor \(e^{x^2/2}\) preserves positivity.
2. For every \(x>0\),\[ f(x)<\frac1x. \]
For \(t>x>0\) one has \(t/x>1\). Hence \[ \begin{aligned} \int_x^\infty e^{-t^2/2}\,dt &< \frac1x \int_x^\infty t e^{-t^2/2}\,dt\\ &= \frac1x e^{-x^2/2}. \end{aligned} \] Multiplying by \(e^{x^2/2}\) gives \[ f(x)<\frac1x. \]
3. Part 1 is\[ \boxed{0<f(x)<\frac1x}. \]
Combine steps 1 and 2.
4. The derivative of \(f\) is\[ f'(x)=xf(x)-1. \]
Differentiate \[ f(x)=e^{x^2/2}\int_x^\infty e^{-t^2/2}\,dt. \] By the product rule and the fundamental theorem of calculus, \[ \begin{aligned} f'(x) &= xe^{x^2/2} \int_x^\infty e^{-t^2/2}\,dt -e^{x^2/2}e^{-x^2/2}\\ &= xf(x)-1. \end{aligned} \]
5. The function \(f\) is strictly decreasing on \((0,\infty)\).
Step 2 gives \(xf(x)<1\) for every \(x>0\). Hence step 4 gives \[ f'(x)=xf(x)-1<0. \] Therefore \(f\) is strictly decreasing.
6. Q.E.D.
Step 3 proves part 1, and step 5 proves part 2.