Help me sir in phase sppace d^3 p=dp_x dp_(y ) dp_z then find ∫


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Question Number 158694 by ilhamdiii last updated on 07/Nov/21

$H e l p m e s i r$ $i n p h a s e s p p a c e d^{3} p = {d p}_{x} {d p}_{y} {d p}_{z} t h e n$ $f i n d$ $\int_{0}^{\infty} P^{2} e^{\frac{p^{2}}{2 m K T}} d^{3} p = . . . .$
	Answered by MrGaster last updated on 06/Jan/25

	$Let x = \frac{p^{2}}{1 m} \Rightarrow p = \sqrt{2 m K T} \times x^{\frac{1}{2}} d p = \sqrt{\frac{m}{2 K p}} \times x^{- 1 / 2} d x$ $\int_{0}^{\infty} p^{2} e^{- p^{2} / (2 m K T)} d^{3} p$ $= \int_{0}^{\infty} {(2 m k T)}^{\frac{3}{2}} \times x^{\frac{3}{2}} \times e^{- x} \times (\frac{\sqrt{m}}{\sqrt{2 K T}}) \times \frac{1}{\sqrt{x}} d x$ $= \frac{(2^{\frac{3}{2}} \times {(m K T)}^{\frac{3}{2}})}{\sqrt{2 π}} \int_{0}^{\infty} x^{\frac{1}{2}} e^{- x} d x$ $\int_{0}^{\infty} x^{1 / 2} e^{- x} d x = Γ (\frac{3}{2}) = \frac{\sqrt{π}}{2}$ $\frac{(2^{3 / 2} \times {(m K T)}^{\frac{3}{2}})}{2 \sqrt{2 π}} \times \sqrt{π}$ $= \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{2 \sqrt{2 π}} \times \sqrt{π}$ $= \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{\sqrt{π}}$ $\int_{0}^{\infty} p^{2} e^{- p^{2} / (2 m K T)} d^{3} p = \frac{\sqrt{2} \times {(m K T)}^{\frac{3}{2}}}{\sqrt{π}}$
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