Any probability density must enclose unit area — that single condition fixes K.

\(\int_{-\infty}^{\infty}p(x)\,dx=1\)

Step 1 — recall the Gaussian integral. The standard result is

\(\int_{-\infty}^{\infty}e^{-ax^{2}}dx=\sqrt{\dfrac{\pi}{a}}\)

and here the exponent is \(-\dfrac{x^{2}}{2}\), so \(a=\dfrac{1}{2}\) and

\(\int_{-\infty}^{\infty}e^{-x^{2}/2}dx=\sqrt{\dfrac{\pi}{1/2}}=\sqrt{2\pi}\)

Step 2 — impose normalisation.

\(K\sqrt{2\pi}=1\qquad\Rightarrow\qquad K=\dfrac{1}{\sqrt{2\pi}}\)

which is option 2, and numerically \(K=0.3989\).

Recognise the distribution. Comparing with the general Gaussian form

\(p(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{(x-\mu)^{2}}{2\sigma^{2}}\right)\)

shows that this is the standard normal distribution, with mean \(\mu=0\) and variance \(\sigma^{2}=1\). The constant \(1/\sqrt{2\pi}\) is therefore not something to derive afresh each time — it is the familiar peak value of the standard normal curve, and recognising it answers the question in one step.

A quick plausibility check on the other options : K must be less than 1 (a density peaking above 1 while spread over an infinite range could not integrate to unity here), which rules out option 1 at 0.798 — it is in fact the correct constant for a half-normal density on \(0\le x\lt\infty\), twice the value found above, and is the trap for anyone who integrates over only half the range.

Why this density dominates communication theory : thermal noise in a resistor is the sum of an enormous number of independent electron motions, and by the central limit theorem any such sum tends to a Gaussian whatever the individual statistics. That is the justification for the AWGN channel model on which every error-probability calculation rests.

Hence, K = 1/√(2π).