Question

If you know plz tell me:
In ideal gas how we get integral of [e^-(βp²/2m) dp] with limit of (-∞ to ∞) is = (2mπ/β)^½

Answer 1

I am too lazy to type the β/2m is i'll let it be k.

If you want a better treatment of this integral, look for gaussian integral.

\[Let\:\int_{-\infty}^{\infty}e^{-ap^{2}}dp=I\]Square both sides
\[\int_{-\infty}^{\infty}\int_{\infty}^{-\infty}e^{-ap^{2}}e^{-aq^{2}}dpdq=I^{2}=\int_{\infty}^{-\infty}\int_{\infty}^{-\infty}e^{-a(p^{2}+q^{2})}dpdq\] Recall that x^2+y^2 represents a circle with radius R, so
\[I^2=\int_{2\pi}^{0}\int_{0}^{\infty}e^{-ar^{2}}rdrd\theta\]
\[I^2=2\pi\int_{0}^{\infty}e^{-ar^{2}}\frac{1}{2}(dr^{2})\]
\[I^2=2\pi\int_{0}^{\infty}e^{-ar^{2}}\frac{1}{2}(dr^{2})\]
\[I^2=0-(-\frac{\pi}{a})\] Thus,\[I=\sqrt{\frac{\pi}{a}}\]