0-dx-1-x-2-4-x-2-16-x-2-64-x-2-

Question Number 122302 by Dwaipayan Shikari last updated on 15/Nov/20

\int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (4 + x^{2}) (16 + x^{2}) (64 + x^{2})}

Answered by TANMAY PANACEA last updated on 15/Nov/20

\frac{1}{48} \int \frac{(64 + x^{2}) - (16 + x^{2})}{(1 + x^{2}) (4 + x^{2}) (16 + x^{2}) (64 + x^{2})} d x

\frac{1}{48} \int \frac{d x}{(1 + x^{2}) (4 + x^{2}) (16 + x^{2})} - \frac{1}{48} \int \frac{d x}{(1 + x^{2}) (4 + x^{2}) (64 + x^{2})}

★

\frac{1}{48 \times 12} \int \frac{d x}{(1 + x^{2}) ((4 + x^{2})} - \frac{1}{48 \times 12} \int \frac{d x}{(1 + x^{2}) (16 + x^{2})}

- \frac{1}{48 \times 60} \int \frac{d x}{(1 + x^{2}) (4 + x^{2})} + \frac{1}{48 \times 60} \int \frac{d x}{(1 + x^{2}) (64 + x^{2})}

\frac{1}{48 \times 12 \times 3} \int \frac{d x}{1 + x^{2}} - \frac{1}{48 \times 12 \times 3} \int \frac{d x}{4 + x^{2}}

- \frac{1}{48 \times 12 \times 15} \int \frac{d x}{1 + x^{2}} + \frac{1}{48 \times 12 \times 15} \int \frac{d x}{16 + x^{2}}

- \frac{1}{48 \times 60 \times 3} \int \frac{d x}{1 + x^{2}} + \frac{1}{48 \times 60 \times 3} \int \frac{d x}{4 + x^{2}}

+ \frac{1}{48 \times 60 \times 63} \int \frac{d x}{1 + x^{2}} - \frac{1}{48 \times 60 \times 63} \int \frac{d x}{64 + x^{2}}

n o w \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} {t a n}^{- 1} (\frac{x}{a}) + c

s o n o w i t e a s y t o s o l v e b u t l e n g t h y

Commented by Dwaipayan Shikari last updated on 15/Nov/20

ধন্যবাদ

Answered by mathmax by abdo last updated on 15/Nov/20

I = \int_{0}^{\infty} \frac{dx}{(x^{2} + 1) (x^{2} + 4) (x^{2} + 16) (x^{2} + 64)} \Rightarrow

2 I = \int_{- \infty}^{+ \infty} \frac{dx}{(x^{2} + 1) (x^{2} + 4) (x^{2} + 16) (x^{2} + 64)} let

φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + 4) (z^{2} + 16) (z^{2} + 64)} \Rightarrow

φ (z) = \frac{1}{(z - i) (z + i) (z - 2 i) (z + 2 i) (z - 4 i) (z + 4 i) (z - 8 i) (z + 8 i)}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, 2 i) + Res (φ, 4 i) + Res (φ, 8 i)}

Res (φ, i) = \frac{1}{2 i (- 1 + 4) (- 1 + 16) (- 1 + 64)} = \frac{1}{2 i . 3.15.63}

Res (φ, 2 i) = \frac{1}{4 i (- 4 + 1) (- 4 + 16) (- 4 + 64)}

= \frac{1}{4 i (- 3) (12) (60)} = - \frac{1}{4 i . 3.12.60}

Res (φ, 4 i) = \frac{1}{8 i (- 16 + 1) (- 16 + 4) (- 16 + 64)}

= \frac{1}{8 i (- 15) (- 12) (48)} = \frac{1}{8 i . 15.12.48}

Res (φ, 8 i) = \frac{1}{16 i (- 64 + 1) (- 64 + 4) (- 64 + 16)}

= - \frac{1}{16 i . 63.60.48} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{1}{2 i . 3.15.63} - \frac{1}{4 i . 3.12.60} + \frac{1}{8 i . 15.12.48} - \frac{1}{16 i . 63.60.48}}

= \frac{π}{3.15.63} - \frac{π}{6.12.60} + \frac{π}{4.15.12.48} - \frac{π}{8.63.60.48} and

I = \frac{1}{2} \int_{- \infty}^{+ \infty} φ (z) dz

Commented by Dwaipayan Shikari last updated on 15/Nov/20

T h a n k i n g y o u

Commented by mathmax by abdo last updated on 15/Nov/20

you are welcome

Answered by MJS_new last updated on 15/Nov/20

\int \frac{d x}{(x^{2} + a) (x^{2} + b) (x^{2} + c) (x^{2} + d)} =

= \frac{1}{(b - a) (c - a) (d - a)} \int \frac{d x}{x^{2} + a} +

+ \frac{1}{(a - b) (c - b) (d - b)} \int \frac{d x}{x^{2} + b} +

+ \frac{1}{(a - c) (b - c) (d - c)} \int \frac{d x}{x^{2} + c} +

+ \frac{1}{(a - d) (b - d) (c - d)} \int \frac{d x}{x^{2} + d}

now {it}^{'} s easy . I get

\int \frac{d x}{(x^{2} + a) (x^{2} + b) (x^{2} + c) (x^{2} + d)} =

= \frac{\arctan x}{2835} - \frac{\arctan \frac{x}{2}}{4320} + \frac{\arctan \frac{x}{4}}{34560} - \frac{\arctan \frac{x}{8}}{1451520} + C

\Rightarrow

answer is \frac{31 π}{414720}

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