# Show $Q\big[\sqrt{\frac{T}{N_{0}}}\big] = \frac{1}{\sqrt{2 \pi N_{0}}} \, \int_{\sqrt{T}}^{\infty} e^{\frac{-x^{2}}{2N_{0}}} \, dx$

I don't understand how:

$Q\big[\sqrt{\frac{T}{N_{0}}}\big] = \frac{1}{\sqrt{2 \pi N_{0}}} \, \int_{\sqrt{T}}^{\infty} e^{\frac{-x^{2}}{2N_{0}}} \, dx$

When the Q function is defined as:

$Q(u) = \frac{1}{ \sqrt{2 \pi}} \, \int_{u}^{\infty} e^{\frac{-x^{2}}{2}} \, dx$

Shouldn't it be that:

$Q(\sqrt{\frac{T}{N_{0}}}) = \frac{1}{ \sqrt{2 \pi}} \, \int_{\sqrt{\frac{T}{N_{0}}}}^{\infty} e^{\frac{-x^{2}}{2}} \, dx$

Are they equivalent? I can't figure out how...I tried plugging both into Maple and even plugging in a range of values and Maple seems to say they aren't equivalent...yet my textbook says they are.

Any ideas?

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Yes, just by plugging in $\sqrt{\frac{T}{N_0}}$ in the expression for $Q$ you should get $$Q\left(\sqrt{\frac{T}{N_0}}\right)=\frac{1}{\sqrt{2\pi}}\int_{\sqrt{\frac{T}{N_0}}}^\infty \exp\left(-\tfrac{1}{2}x^2\right)\,\mathrm dx.$$ But it turns out that by a change of variables you can obtain the other expression. Let us substitute $y=x\sqrt{N_0}$ or equivalently $x=\frac{y}{\sqrt{N_0}}$ in the integral. Note that $$\frac{\mathrm dy}{\mathrm dx}=\sqrt{N_0}\iff\mathrm dx=\frac{1}{\sqrt{N_0}}\,\mathrm dy$$ and if $x=\sqrt{\frac{T}{N_0}}$ then $y=\sqrt{T}$ and if $x=\infty$ then also $y=\infty$ (this is to see what the upper and lower limit of the integral should be after changing the variable). Thus \begin{align} \frac{1}{\sqrt{2\pi}}\int_{\sqrt{\frac{T}{N_0}}}^\infty \exp\left(-\tfrac{1}{2}x^2\right)\,\mathrm dx&=\frac{1}{\sqrt{2\pi}}\int_\sqrt{T}^\infty\exp\left(-\frac{1}{2}\left(\frac{y}{\sqrt{N_0}}\right)^2\right)\cdot\frac{1}{\sqrt{N_0}}\,\mathrm dy\\ &=\frac{1}{\sqrt{2\pi N_0}}\int_\sqrt{T}^\infty\exp\left(-\frac{y^2}{2N_0}\right)\,\mathrm dy. \end{align}

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You should correct those $\frac{1}{2\pi}$'s to $\frac{1}{\sqrt{2\pi}}$'s. – Dilip Sarwate Mar 17 '13 at 16:02
@DilipSarwate: Yes, thank you. – Stefan Hansen Mar 17 '13 at 16:06
Thanks! This is what I was looking for! – TheDude Mar 18 '13 at 18:00

Hint: let $x=v/\sqrt{N_0}$ in your integral.

\begin{align}Q(\sqrt{\frac{T}{N_{0}}}) &= \frac{1}{ \sqrt{2 \pi }} \, \int_{\sqrt{\frac{T}{N_{0}}}}^{\infty} dx \:e^{\frac{-x^{2}}{2}} \\ &= \frac{1}{ \sqrt{2 \pi}} \frac{1}{\sqrt{N_0}} \, \int_{\sqrt{\frac{T}{N_{0}}} \sqrt{N_0}}^{\infty} dv \: e^{\frac{-v^{2}}{2 N_0}} \\ &= \frac{1}{ \sqrt{2 \pi N_0}} \, \int_{\sqrt{T}}^{\infty} dx \:e^{\frac{-x^{2}}{2 N_0}}\end{align}

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I'm not sure how that helps... Perhaps I just don't know how to evaluate that type of integral. A more explanatory/basic answer is what is needed here. – TheDude Mar 17 '13 at 4:02