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

∫_0 ^∞ (dx/((1+x^2 )(4+x^2 )(16+x^2 )(64+x^2 )))

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{16}+{x}^{\mathrm{2}} \right)\left(\mathrm{64}+{x}^{\mathrm{2}} \right)} \\ $$

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

(1/(48))∫(((64+x^2 )−(16+x^2 ))/((1+x^2 )(4+x^2 )(16+x^2 )(64+x^2 )))dx (1/(48))∫(dx/((1+x^2 )(4+x^2 )(16+x^2 )))−(1/(48))∫(dx/((1+x^2 )(4+x^2 )(64+x^2 ))) ★ (1/(48×12))∫(dx/((1+x^2 )((4+x^2 )))−(1/(48×12))∫(dx/((1+x^2 )(16+x^2 ))) −(1/(48×60))∫(dx/((1+x^2 )(4+x^2 )))+(1/(48×60))∫(dx/((1+x^2 )(64+x^2 ))) ■ (1/(48×12×3))∫(dx/(1+x^2 ))−(1/(48×12×3))∫(dx/(4+x^2 )) −(1/(48×12×15))∫(dx/(1+x^2 ))+(1/(48×12×15))∫(dx/(16+x^2 )) −(1/(48×60×3))∫(dx/(1+x^2 ))+(1/(48×60×3))∫(dx/(4+x^2 )) +(1/(48×60×63))∫(dx/(1+x^2 ))−(1/(48×60×63))∫(dx/(64+x^2 )) now ∫(dx/(x^2 +a^2 ))=(1/a)tan^(−1) ((x/a))+c so now it easy tosolve but lengthy

$$\frac{\mathrm{1}}{\mathrm{48}}\int\frac{\left(\mathrm{64}+{x}^{\mathrm{2}} \right)−\left(\mathrm{16}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{16}+{x}^{\mathrm{2}} \right)\left(\mathrm{64}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{48}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{16}+{x}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{48}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{64}+{x}^{\mathrm{2}} \right)} \\ $$$$\bigstar \\ $$$$\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\right.}−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{16}+{x}^{\mathrm{2}} \right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}}\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{64}+{x}^{\mathrm{2}} \right)} \\ $$$$\blacksquare \\ $$$$\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}×\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}×\mathrm{3}}\int\frac{{dx}}{\mathrm{4}+{x}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}×\mathrm{15}}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{48}×\mathrm{12}×\mathrm{15}}\int\frac{{dx}}{\mathrm{16}+{x}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}×\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}×\mathrm{3}}\int\frac{{dx}}{\mathrm{4}+{x}^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}×\mathrm{63}}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{48}×\mathrm{60}×\mathrm{63}}\int\frac{{dx}}{\mathrm{64}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\boldsymbol{{now}}\:\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)+{c} \\ $$$${so}\:{now}\:{it}\:{easy}\:{tosolve}\:{but}\:{lengthy} \\ $$

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

ধন্যবাদ

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

I =∫_0 ^∞ (dx/((x^2 +1)(x^2 +4)(x^2 +16)(x^2 +64))) ⇒ 2I =∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +4)(x^2 +16)(x^2 +64))) let ϕ(z)=(1/((z^2 +1)(z^2 +4)(z^2 +16)(z^2 +64))) ⇒ ϕ(z)=(1/((z−i)(z+i)(z−2i)(z+2i)(z−4i)(z+4i)(z−8i)(z+8i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i)+Res(ϕ,2i) +Res(ϕ,4i)+Res(ϕ ,8i)} Res(ϕ,i)=(1/(2i(−1+4)(−1+16)(−1+64)))=(1/(2i.3.15.63)) Res(ϕ,2i) =(1/(4i(−4+1)(−4+16)(−4+64))) =(1/(4i(−3)(12)(60))) =−(1/(4i.3.12.60)) Res(ϕ,4i) =(1/(8i(−16+1)(−16+4)(−16+64))) =(1/(8i(−15)(−12)(48))) =(1/(8i.15.12.48)) Res(ϕ,8i)=(1/(16i(−64+1)(−64+4)(−64+16))) =−(1/(16i.63.60.48)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(2i.3.15.63))−(1/(4i.3.12.60))+(1/(8i.15.12.48))−(1/(16i.63.60.48))} =(π/(3.15.63)) −(π/(6.12.60)) +(π/(4.15.12.48))−(π/(8.63.60.48)) and I =(1/2)∫_(−∞) ^(+∞) ϕ(z)dz

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{16}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{64}\right)}\:\Rightarrow \\ $$$$\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{16}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{64}\right)}\:\mathrm{let}\: \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{16}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{64}\right)}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{2i}\right)\left(\mathrm{z}−\mathrm{4i}\right)\left(\mathrm{z}+\mathrm{4i}\right)\left(\mathrm{z}−\mathrm{8i}\right)\left(\mathrm{z}+\mathrm{8i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\mathrm{Res}\left(\varphi,\mathrm{i}\right)+\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\:+\mathrm{Res}\left(\varphi,\mathrm{4i}\right)+\mathrm{Res}\left(\varphi\:,\mathrm{8i}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\frac{\mathrm{1}}{\mathrm{2i}\left(−\mathrm{1}+\mathrm{4}\right)\left(−\mathrm{1}+\mathrm{16}\right)\left(−\mathrm{1}+\mathrm{64}\right)}=\frac{\mathrm{1}}{\mathrm{2i}.\mathrm{3}.\mathrm{15}.\mathrm{63}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\:=\frac{\mathrm{1}}{\mathrm{4i}\left(−\mathrm{4}+\mathrm{1}\right)\left(−\mathrm{4}+\mathrm{16}\right)\left(−\mathrm{4}+\mathrm{64}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4i}\left(−\mathrm{3}\right)\left(\mathrm{12}\right)\left(\mathrm{60}\right)}\:=−\frac{\mathrm{1}}{\mathrm{4i}.\mathrm{3}.\mathrm{12}.\mathrm{60}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{4i}\right)\:=\frac{\mathrm{1}}{\mathrm{8i}\left(−\mathrm{16}+\mathrm{1}\right)\left(−\mathrm{16}+\mathrm{4}\right)\left(−\mathrm{16}+\mathrm{64}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8i}\left(−\mathrm{15}\right)\left(−\mathrm{12}\right)\left(\mathrm{48}\right)}\:=\frac{\mathrm{1}}{\mathrm{8i}.\mathrm{15}.\mathrm{12}.\mathrm{48}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{8i}\right)=\frac{\mathrm{1}}{\mathrm{16i}\left(−\mathrm{64}+\mathrm{1}\right)\left(−\mathrm{64}+\mathrm{4}\right)\left(−\mathrm{64}+\mathrm{16}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{16i}.\mathrm{63}.\mathrm{60}.\mathrm{48}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2i}.\mathrm{3}.\mathrm{15}.\mathrm{63}}−\frac{\mathrm{1}}{\mathrm{4i}.\mathrm{3}.\mathrm{12}.\mathrm{60}}+\frac{\mathrm{1}}{\mathrm{8i}.\mathrm{15}.\mathrm{12}.\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{16i}.\mathrm{63}.\mathrm{60}.\mathrm{48}}\right\} \\ $$$$=\frac{\pi}{\mathrm{3}.\mathrm{15}.\mathrm{63}}\:−\frac{\pi}{\mathrm{6}.\mathrm{12}.\mathrm{60}}\:+\frac{\pi}{\mathrm{4}.\mathrm{15}.\mathrm{12}.\mathrm{48}}−\frac{\pi}{\mathrm{8}.\mathrm{63}.\mathrm{60}.\mathrm{48}}\:\:\mathrm{and}\: \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\: \\ $$$$ \\ $$

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

Thanking you

$${Thanking}\:{you} \\ $$

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

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Answered by MJS_new last updated on 15/Nov/20

∫(dx/((x^2 +a)(x^2 +b)(x^2 +c)(x^2 +d)))= =(1/((b−a)(c−a)(d−a)))∫(dx/(x^2 +a))+ +(1/((a−b)(c−b)(d−b)))∫(dx/(x^2 +b))+ +(1/((a−c)(b−c)(d−c)))∫(dx/(x^2 +c))+ +(1/((a−d)(b−d)(c−d)))∫(dx/(x^2 +d)) now it′s easy. I get ∫(dx/((x^2 +a)(x^2 +b)(x^2 +c)(x^2 +d)))= =((arctan x)/(2835))−((arctan (x/2))/(4320))+((arctan (x/4))/(34560))−((arctan (x/8))/(1451520))+C ⇒ answer is ((31π)/(414720))

$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}\right)\left({x}^{\mathrm{2}} +{b}\right)\left({x}^{\mathrm{2}} +{c}\right)\left({x}^{\mathrm{2}} +{d}\right)}= \\ $$$$=\frac{\mathrm{1}}{\left({b}−{a}\right)\left({c}−{a}\right)\left({d}−{a}\right)}\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}}+ \\ $$$$+\frac{\mathrm{1}}{\left({a}−{b}\right)\left({c}−{b}\right)\left({d}−{b}\right)}\int\frac{{dx}}{{x}^{\mathrm{2}} +{b}}+ \\ $$$$+\frac{\mathrm{1}}{\left({a}−{c}\right)\left({b}−{c}\right)\left({d}−{c}\right)}\int\frac{{dx}}{{x}^{\mathrm{2}} +{c}}+ \\ $$$$+\frac{\mathrm{1}}{\left({a}−{d}\right)\left({b}−{d}\right)\left({c}−{d}\right)}\int\frac{{dx}}{{x}^{\mathrm{2}} +{d}} \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}.\:\mathrm{I}\:\mathrm{get} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}\right)\left({x}^{\mathrm{2}} +{b}\right)\left({x}^{\mathrm{2}} +{c}\right)\left({x}^{\mathrm{2}} +{d}\right)}= \\ $$$$=\frac{\mathrm{arctan}\:{x}}{\mathrm{2835}}−\frac{\mathrm{arctan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{4320}}+\frac{\mathrm{arctan}\:\frac{{x}}{\mathrm{4}}}{\mathrm{34560}}−\frac{\mathrm{arctan}\:\frac{{x}}{\mathrm{8}}}{\mathrm{1451520}}+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{31}\pi}{\mathrm{414720}} \\ $$

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