دالة كاما gamma function (شرح بسيط)

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$Q1)\int_{0}^{\infty }\sqrt{x}e^{-x^{3}}dx$

$y=x^{3} , y^{\frac{1}{3}}=x , dx=\frac{1}{3}y^{\frac{-2}{3}}$

$=\frac{1}{3}\int_{0}^{\infty }y^{\frac{1}{6}}e^{-y}y^{\frac{-2}{3}}dy$

$=\frac{1}{3}\int_{0}^{\infty }y^{(-\frac{1}{2})}e^{-y}dy$

$=\frac{1}{3}\int_{0}^{\infty }y^{(\frac{1}{2}-1)}e^{-y}dy$

$=\frac{1}{3}\Gamma (\frac{1}{2})$

$Q2)\int_{0}^{\infty }3^{-4z^{2}}dz$

$e^{-y}=3^{-4z^{2}},y=4z^{2}ln3,z=\frac{\sqrt{y}}{2\sqrt{ln3}} ,dz=\frac{dy}{4\sqrt{ln3}\sqrt{y}}$

$=(\frac{1}{4\sqrt{ln3}})\int_{0}^{\infty }y^{-\frac{1}{2}}e^{-y}dy$

$=(\frac{1}{4\sqrt{ln3}})\Gamma (\frac{1}{2})$

$Q3)\int_{0}^{1}\frac{1}{\sqrt{-ln(t)}}dt$

$y=-lnt,-y=lnt,e^{-y}=t,dt=-e^{-y}dy$

$=\int_{0}^{\infty }y^{-\frac{1}{2}}e^{-y}dy=\Gamma (\frac{1}{2})$

حل آخر

$Q3)\int_{0}^{1}\frac{1}{\sqrt{-lnt}}dt$

$y=\sqrt{-lnt},y^{2}=-lnt,e^{y^{2}}=\frac{1}{t},t=e^{-y^{2}},dt=-2ye^{-y^{2}}dy$

$t=0\Rightarrow y=\infty ,t=1\Rightarrow y=0$

$=\int_{0}^{\infty }\frac{1}{y}2ye^{-y^{2}}dy=\int_{0}^{\infty }2e^{-y^{2}}dy$

$z=y^{2},\sqrt{z}=y,dy=\frac{1}{2\sqrt{z}}dz$

$=\int_{0}^{\infty }z^{-\frac{1}{2}}e^{-z}dz=\Gamma (\frac{1}{2})$

$Q4)\int_{0}^{1}\sqrt[3]{ln(\frac{1}{y})}dy$

$z=ln(\frac{1}{y}),e^{z}=\frac{1}{y},y=e^{-z},dy=-e^{-z}dz$

$=\int_{0}^{\infty }z^{\frac{1}{3}}e^{-z}dz=\Gamma (\frac{4}{3})$

حل آخر


$Q4)\int_{0}^{1}\sqrt[3]{ln(\frac{1}{y})}dy$

$z=\sqrt[3]{ln(\frac{1}{y})},z^{3}=ln(\frac{1}{y}),e^{z^{3}}= \frac{1}{y},y=e^{-z^{3}},dy=-3z^{2}e^{-z^{3}}$

$y=0\Rightarrow z=\infty ,y=1\Rightarrow z=0$

$=\int_{0}^{\infty }3z^{3}e^{-z^{3}}dz$

$z^{3}=x,z=\sqrt[3]{x},dz=\frac{1}{3}x^{\frac{-2}{3}}dx$

$=\int_{0}^{\infty }x^{\frac{1}{3}}e^{-x}dx=\Gamma (\frac{4}{3})$