Hi! Does anybody know how one can proof that $\displaystyle \frac{c}{1+c^2}\cdot exp(-\frac{c^2}{2}) \leq \int_c^\infty exp(-\frac{z^2}{2}) dz$ for $\displaystyle c>0$

I am thankful for any ideas.

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- May 16th 2009, 10:22 PMgammafunctioninequation with an integral
Hi! Does anybody know how one can proof that $\displaystyle \frac{c}{1+c^2}\cdot exp(-\frac{c^2}{2}) \leq \int_c^\infty exp(-\frac{z^2}{2}) dz$ for $\displaystyle c>0$

I am thankful for any ideas. - May 17th 2009, 01:13 AMNonCommAlg
let $\displaystyle z=c\sqrt{2x+1}.$ then $\displaystyle I=\int_c^{\infty} \exp \left(\frac{-z^2}{2} \right) \ dz=c \exp \left(\frac{-c^2}{2} \right) \int_0^{\infty} \frac{e^{-c^2x}}{\sqrt{2x+1}} \ dx.$ but we know that $\displaystyle e^a \geq 1+a,$ for any $\displaystyle a \geq 0.$ thus $\displaystyle \frac{1}{\sqrt{2x+1}} \geq e^{-x}$ and hence:

$\displaystyle I \geq c \exp \left(\frac{-c^2}{2} \right) \int_0^{\infty} e^{-(1+c^2)x} \ dx=\frac{c}{1+c^2} \exp \left(\frac{-c^2}{2} \right).$

it was a little tricky, wasn't it? (Wink) - May 17th 2009, 01:41 AMgammafunction
Tricky and wonderful (Rofl). What a nice proof. Thank you a thousand times.