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Question-115438




Question Number 115438 by A8;15: last updated on 25/Sep/20
Answered by Olaf last updated on 26/Sep/20
the function is odd  and the domain is symmetrical  so the result is 0.  Sorry I′m wrong.  see mister abdo.
$${the}\:{function}\:{is}\:{odd} \\ $$$${and}\:{the}\:{domain}\:{is}\:{symmetrical} \\ $$$${so}\:{the}\:{result}\:{is}\:\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{Sorry}}\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{wrong}}. \\ $$$$\boldsymbol{\mathrm{see}}\:\boldsymbol{\mathrm{mister}}\:\boldsymbol{\mathrm{abdo}}. \\ $$
Commented by mathmax by abdo last updated on 26/Sep/20
here {x^3 } mean x^3 −[x^3 ]...!
$$\mathrm{here}\:\left\{\mathrm{x}^{\mathrm{3}} \right\}\:\mathrm{mean}\:\mathrm{x}^{\mathrm{3}} −\left[\mathrm{x}^{\mathrm{3}} \right]…! \\ $$
Commented by Olaf last updated on 26/Sep/20
sorry mister !  I′m french. In my country we don′t  use the same notations. Sorry !
$$\mathrm{sorry}\:\mathrm{mister}\:! \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{french}.\:\mathrm{In}\:\mathrm{my}\:\mathrm{country}\:\mathrm{we}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{use}\:\mathrm{the}\:\mathrm{same}\:\mathrm{notations}.\:\mathrm{Sorry}\:! \\ $$
Commented by Bird last updated on 26/Sep/20
nevermind
$${nevermind} \\ $$
Answered by mathmax by abdo last updated on 26/Sep/20
I =∫_(−1) ^1  (({x^3 }(x^4 +1))/(x^6  +1))dx ⇒I =∫_(−1) ^1  (((x^3 −[x^3 ](x^4  +1))/(x^6  +1))dx  =∫_(−1) ^(1 )  ((x^3 (x^4 +1))/(x^6 +1))dx(→=0  odd function) −∫_(−1) ^1  (([x^3 ](x^4 +1))/(x^6  +1))dx  we haveA= ∫_(−1) ^1  (([x^3 ](x^4 +1))/(x^6  +1))dx  =_(x=−t)   ∫_(−1) ^1  (([−t^3 ](t^4 +1))/(t^6  +1))dt ⇒  2A =∫_(−1) ^1  ((([x^3 ]+[−x^3 ])(x^4  +1))/(x^6 +1))dx =∫_(−1) ^0  ((([x^3 ]+[−x^3 ])(x^4  +1))/(x^6  +1))dx(x=−t)  +∫_0 ^1  ((([x^3 ]+[−x^3 ])(x^4 +1))/(x^6 +1))dx  =2∫_0 ^1   ((([x^3 ]+[−x^3 ])(x^4  +1))/(x^6  +1))dx =−2 ∫_0 ^1  ((1+x^4 )/(1+x^6 )) dx  =−2 ∫_0 ^1 (1+x^4 )Σ_(n=0) ^∞  (−1)^n  x^(6n)   =−2 Σ_(n=0) ^∞  (−1)^n ∫_0 ^1 (x^(6n)  +x^(6n+4) )dx  =−2 Σ_(n=0) ^∞  (((−1)^n )/(6n+1))−2 Σ_(n=0) ^∞  (((−1)^n )/(6n+5))  ...be continued  we can calculate ∫_0 ^1  ((1+x^4 )/(1+x^6 ))dx by elementary functions...
$$\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left\{\mathrm{x}^{\mathrm{3}} \right\}\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left(\mathrm{x}^{\mathrm{3}} −\left[\mathrm{x}^{\mathrm{3}} \right]\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)\right.}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}\:} \:\frac{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\mathrm{dx}\left(\rightarrow=\mathrm{0}\:\:\mathrm{odd}\:\mathrm{function}\right)\:−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left[\mathrm{x}^{\mathrm{3}} \right]\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{haveA}=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left[\mathrm{x}^{\mathrm{3}} \right]\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx}\:\:=_{\mathrm{x}=−\mathrm{t}} \:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left[−\mathrm{t}^{\mathrm{3}} \right]\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{t}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{2A}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\left(\left[\mathrm{x}^{\mathrm{3}} \right]+\left[−\mathrm{x}^{\mathrm{3}} \right]\right)\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\mathrm{dx}\:=\int_{−\mathrm{1}} ^{\mathrm{0}} \:\frac{\left(\left[\mathrm{x}^{\mathrm{3}} \right]+\left[−\mathrm{x}^{\mathrm{3}} \right]\right)\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx}\left(\mathrm{x}=−\mathrm{t}\right) \\ $$$$+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\left[\mathrm{x}^{\mathrm{3}} \right]+\left[−\mathrm{x}^{\mathrm{3}} \right]\right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} +\mathrm{1}}\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\left[\mathrm{x}^{\mathrm{3}} \right]+\left[−\mathrm{x}^{\mathrm{3}} \right]\right)\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\:\mathrm{dx} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{6n}} \:\:=−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{6n}} \:+\mathrm{x}^{\mathrm{6n}+\mathrm{4}} \right)\mathrm{dx} \\ $$$$=−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{6n}+\mathrm{1}}−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{6n}+\mathrm{5}}\:\:…\mathrm{be}\:\mathrm{continued} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx}\:\mathrm{by}\:\mathrm{elementary}\:\mathrm{functions}… \\ $$

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