Definite Integration Test - 2

Similar solution with @Karan Chatrath's, but I promise with more details.

$\begin{aligned} I & = \int_0^\infty \frac 1{1+x^{16}} dx & \small \color{#3D99F6} x^8 = \tan \theta \implies 8x^7 dx = \sec^2 \theta \ d\theta \\ & = \frac 18 \int_0^\frac \pi 2 \sin^{-\frac 78} \theta \cos^\frac 78 \theta\ d\theta & \small \color{#3D99F6} \text{Beta function } B(m,n) = 2 \int_0^\frac \pi 2 \sin^{2m-1} x \cos^{2n-1} x \ dx \\ & = \frac {B\left(\frac 1{16}, \frac {15}{16} \right)}{16} & \small \color{#3D99F6} B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \text{. where }\Gamma (\cdot) \text{ is gamma function.} \\ & = \frac {\color{#3D99F6}\Gamma \left(\frac 1{16}\right)\Gamma \left(\frac {15}{16} \right)}{16\color{#D61F06}\Gamma (1)} & \small \color{#3D99F6} \Gamma(z) \Gamma(1-z) = \frac \pi{\sin \pi z} \\ & = \frac {\color{#3D99F6}\pi \csc \frac \pi{16}}{16\times\color{#D61F06}0!} & \small \color{#D61F06} \Gamma (n) = (n-1)! \\ & = \frac 1{16}\csc \frac \pi{16} \end{aligned}$

Therefore, $A=16$ . Since the volume of revolution of ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ along the $x$ -axis is given by $V = \dfrac 43 \pi ab^2$ , where $a= A = 16$ and $b=A-3= 13$ , then $\lfloor V \rfloor = \left \lfloor \dfrac 43\pi (16)(13^2) \right \rfloor = \boxed{11326}$ .

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References:

• Beta function
• Gamma function
• Volume of ellipsoid

Consider the integral:

$I =\int_{0}^{\infty}\frac{1}{1+x^{16}}dx$

Taking $x^8 = \tan(p)$

Implies: $8x^7 dx = \sec^2(p) dp$

Plugging this back into the integral transforms it to:

$I = \frac{1}{8}\int_{0}^{\frac{\pi}{2}}(\cos(p))^{\frac{7}{8}}(\sin(p))^{-\frac{7}{8}}dp$

This can be easily solved by using the concept of Gamma functions.

$I = \frac{1}{8}\left(\frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}\right)$

Where $m = \frac{7}{8}$ and $n= -\frac{7}{8}$ .

Substituting $m$ and $n$ gives:

$I = \frac{1}{16} \Gamma\left(\frac{15}{16}\right) \Gamma\left(\frac{1}{16}\right)$

Simplifying this expression involves the use of Euler's reflection formula which is:

$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}$

Where $z<1$ . This leads to:

$I = \frac{\pi}{16} \cosec\left(\frac{\pi}{16}\right)$

Hence $\boxed{ A = 16}$

Solving this integral has involved the use of some standard results. The derivation of these results are interesting exercises by themselves. I suggest that the reader looks up relevant concepts. The second part of this problem is relatively simple. One needs to solve the following integral to obtain the answer. I am refraining from describing the process of computing volume of revolution.

$V = \int_{-16}^{16} \left({13}^2\right)\pi\left(1-\frac{x^2}{256}\right)dx$