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MAT334--Lectures & Home Assignments / Re: 2.6 Q22
« on: November 22, 2018, 02:41:00 PM »
For Question(b),$$ \int_0^\infty \frac{xd_x}{x^4+16} = \frac{1}{16} \int_0^\infty \frac{xd_x}{\frac{x^4}{16}+1}$$
Change variable $x/2$ to $t$ and apply 8, then it becomes
$$2\int_0^\infty \frac{2td_t}{1+ t^4} = 4\int_0^\infty \frac{td_t}{1+ t^4}$$
$\beta = 4, \alpha\beta-1 = 1$, so $\alpha = \frac{1}{2}$
$$\int_0^\infty \frac{xd_x}{\frac{x^4}{16}+1}=\frac{4}{4}\frac{\pi}{\sin(\frac{\pi}{2})}=\pi$$
so$$\int_0^\infty \frac{xd_x}{x^4+16} = \frac{\pi}{16} $$
Question(c) is the same following the same logic, with $0\leq\gamma\lt\beta-1$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta}$$
$$\alpha\beta-1=\gamma$$
$$\alpha=\frac{\gamma+1}{\beta}, \frac{1}{\beta}\le\frac{\gamma+1}{\beta}\lt1$$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta} = \frac{\pi}{\beta\sin(\frac{(\gamma+1)\pi}{\beta})}$$
Change variable $x/2$ to $t$ and apply 8, then it becomes
$$2\int_0^\infty \frac{2td_t}{1+ t^4} = 4\int_0^\infty \frac{td_t}{1+ t^4}$$
$\beta = 4, \alpha\beta-1 = 1$, so $\alpha = \frac{1}{2}$
$$\int_0^\infty \frac{xd_x}{\frac{x^4}{16}+1}=\frac{4}{4}\frac{\pi}{\sin(\frac{\pi}{2})}=\pi$$
so$$\int_0^\infty \frac{xd_x}{x^4+16} = \frac{\pi}{16} $$
Question(c) is the same following the same logic, with $0\leq\gamma\lt\beta-1$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta}$$
$$\alpha\beta-1=\gamma$$
$$\alpha=\frac{\gamma+1}{\beta}, \frac{1}{\beta}\le\frac{\gamma+1}{\beta}\lt1$$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta} = \frac{\pi}{\beta\sin(\frac{(\gamma+1)\pi}{\beta})}$$