Rayleigh distribution Density function $\rightarrow f_{x}(x)=\lambda t \cdot \exp \left(\frac{-\lambda t^{2}}{2}\right)$ expected value $\rightarrow E(t)=\int_{0}^{\infty} \lambda t^{2} \cdot \operatorname{eap}\left(\frac{-\lambda t^{2}}{2}\right) d t$ by replacing u insted of $t^{2} \rightarrow 2 t d t=d u \Rightarrow d t=\frac{d u}{2 \sqrt{u}}$ $\int_{0}^{\infty} \frac{1}{2} \cdot u^{1 / 2} e^{-\lambda / 2 u} d u \rightarrow E(t)=\sqrt{\frac{\pi}{2 \lambda}}$ $\operatorname{var}(t)=E\left(t^{2}\right)-(E(t))^{2}=\frac{2}{\lambda}\left(1-\frac{x}{4}\right)$ $E\left(t^{2}\right)=\int_{0}^{\infty} \lambda t^{3} \cdot \exp \left(\frac{-\lambda t^{2}}{2}\right)=?$
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I need step by step solving of this integral .
Let $u=t^2$, $du=2t\,dt$, $dv=\lambda te^{-\lambda t^2/2}dv$, $v=-e^{-\lambda t^2/2}$ then $$\begin{align}\int_0^{\infty}\lambda t^3e^{-\lambda t^2/2}dt&=\int_0^{\infty}u\,dv=\left.uv\right|_0^{\infty}-\int_0^{\infty}v\,du\\ &=\left.-t^2e^{-\lambda t^2/2}\right|_0^{\infty}+2\int_0^{\infty}te^{-\lambda t^2/2}dt\\ &=\left.-\frac2{\lambda}e^{-\lambda t^2/2}\right|_0^{\infty}=\frac2{\lambda}\end{align}$$