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Question-96148




Question Number 96148 by Farruxjano last updated on 30/May/20
Commented by Farruxjano last updated on 30/May/20
Please i need the solution  :(
$$\mathrm{Please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{solution}\:\::\left(\right. \\ $$
Commented by john santu last updated on 30/May/20
∫_0 ^∞  (dx/(1+x^(2n) )) = ((πi)/(2n)) . (((e^((πi)/(2n)) ))/((e^((πi)/n) −1)))   = ((πi)/(2n)) . (e^((πi)/(2n)) /(e^((πi)/(2n))  (e^((πi)/(2n)) −e^(−((πi)/(2n))) )))  = (π/(2n)). (i/(e^((πi)/(2n)) −e^(−((πi)/(2n))) )) = (π/(2n)). (1/(sin ((π/(2n)))))  = (π/(2n.sin ((π/(2n))))) .
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }\:=\:\frac{\pi{i}}{\mathrm{2}{n}}\:.\:\frac{\left({e}^{\frac{\pi{i}}{\mathrm{2}{n}}} \right)}{\left({e}^{\frac{\pi{i}}{{n}}} −\mathrm{1}\right)}\: \\ $$$$=\:\frac{\pi{i}}{\mathrm{2}{n}}\:.\:\frac{{e}^{\frac{\pi{i}}{\mathrm{2}{n}}} }{{e}^{\frac{\pi{i}}{\mathrm{2}{n}}} \:\left({e}^{\frac{\pi{i}}{\mathrm{2}{n}}} −{e}^{−\frac{\pi{i}}{\mathrm{2}{n}}} \right)} \\ $$$$=\:\frac{\pi}{\mathrm{2n}}.\:\frac{{i}}{{e}^{\frac{\pi{i}}{\mathrm{2}{n}}} −{e}^{−\frac{\pi{i}}{\mathrm{2}{n}}} }\:=\:\frac{\pi}{\mathrm{2n}}.\:\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$$$=\:\frac{\pi}{\mathrm{2n}.\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2n}}\right)}\:.\: \\ $$
Commented by mr W last updated on 30/May/20
everyone knows: when you post a  question here, you want to get answer.  you don′t need to repeat again and  again that you need a solution.    but you should also know: nobody owes  you anything! you are not the god  and all others mustn′t dance to your  tune!    besides, it disturbs the order of the  forum when you repeat the same question  by opening alot of new threads.
$${everyone}\:{knows}:\:{when}\:{you}\:{post}\:{a} \\ $$$${question}\:{here},\:{you}\:{want}\:{to}\:{get}\:{answer}. \\ $$$${you}\:{don}'{t}\:{need}\:{to}\:{repeat}\:{again}\:{and} \\ $$$${again}\:{that}\:{you}\:{need}\:{a}\:{solution}. \\ $$$$ \\ $$$${but}\:{you}\:{should}\:{also}\:{know}:\:{nobody}\:{owes} \\ $$$${you}\:{anything}!\:{you}\:{are}\:{not}\:{the}\:{god} \\ $$$${and}\:{all}\:{others}\:{mustn}'{t}\:{dance}\:{to}\:{your} \\ $$$${tune}! \\ $$$$ \\ $$$${besides},\:{it}\:{disturbs}\:{the}\:{order}\:{of}\:{the} \\ $$$${forum}\:{when}\:{you}\:{repeat}\:{the}\:{same}\:{question} \\ $$$${by}\:{opening}\:{alot}\:{of}\:{new}\:{threads}. \\ $$
Commented by mr W last updated on 30/May/20
that′s why i red flagged this post of  you.
$${that}'{s}\:{why}\:{i}\:{red}\:{flagged}\:{this}\:{post}\:{of} \\ $$$${you}. \\ $$
Commented by mr W last updated on 30/May/20
i said i have red flagged this post and i  also said why. who turned the red  flag to “like” again and why?
$${i}\:{said}\:{i}\:{have}\:{red}\:{flagged}\:{this}\:{post}\:{and}\:{i} \\ $$$${also}\:{said}\:{why}.\:{who}\:{turned}\:{the}\:{red} \\ $$$${flag}\:{to}\:“{like}''\:{again}\:{and}\:{why}? \\ $$
Commented by Tinku Tara last updated on 30/May/20
Hi Farruxjano  I see this is just probably first post  from you. This forum to enable   people to learn from each other.  It is ok to write part answers  to show ur attempt.    Just dont create too many threads  for the same question.
$$\mathrm{Hi}\:\mathrm{Farruxjano} \\ $$$$\mathrm{I}\:\mathrm{see}\:\mathrm{this}\:\mathrm{is}\:\mathrm{just}\:\mathrm{probably}\:\mathrm{first}\:\mathrm{post} \\ $$$$\mathrm{from}\:\mathrm{you}.\:\mathrm{This}\:\mathrm{forum}\:\mathrm{to}\:\mathrm{enable}\: \\ $$$$\mathrm{people}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{to}\:\mathrm{write}\:\mathrm{part}\:\mathrm{answers} \\ $$$$\mathrm{to}\:\mathrm{show}\:\mathrm{ur}\:\mathrm{attempt}. \\ $$$$ \\ $$$$\mathrm{Just}\:\mathrm{dont}\:\mathrm{create}\:\mathrm{too}\:\mathrm{many}\:\mathrm{threads} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}. \\ $$
Commented by Farruxjano last updated on 30/May/20
I′m terribly sorry, ok, I′ve understood!
$$\mathrm{I}'\mathrm{m}\:\mathrm{terribly}\:\mathrm{sorry},\:\mathrm{ok},\:\mathrm{I}'\mathrm{ve}\:\mathrm{understood}! \\ $$
Commented by mr W last updated on 30/May/20
i appreciate it very much that you  showed your understanding and said  sorry.
$${i}\:{appreciate}\:{it}\:{very}\:{much}\:{that}\:{you} \\ $$$${showed}\:{your}\:{understanding}\:{and}\:{said} \\ $$$${sorry}. \\ $$
Commented by Rasheed.Sindhi last updated on 30/May/20
Tinku Tara,The developer  I,as an experiment marked the  above post as red_flaged when  it was already red_flagged by  mr W sir. The result was ′a like  is added′ to the post and the post  was clear of red_flag!!!(Sorry  mr W)This  behavior of the app is mathematically  correct where negative of negative  is always positive.....  But I, at the moment, not thinking   mathematically rigister it   as a complaint!!!
$$\mathcal{T}{inku}\:\mathcal{T}{ara},\mathcal{T}{he}\:{developer} \\ $$$${I},{as}\:{an}\:{experiment}\:{marked}\:{the} \\ $$$${above}\:{post}\:{as}\:{red\_flaged}\:{when} \\ $$$${it}\:{was}\:{already}\:{red\_flagged}\:{by} \\ $$$${mr}\:{W}\:{sir}.\:{The}\:{result}\:{was}\:'{a}\:{like} \\ $$$${is}\:{added}'\:{to}\:{the}\:{post}\:{and}\:{the}\:{post} \\ $$$${was}\:{clear}\:{of}\:{red\_flag}!!!\left({Sorry}\:\:{mr}\:{W}\right){This} \\ $$$${behavior}\:{of}\:{the}\:{app}\:{is}\:{mathematically} \\ $$$${correct}\:{where}\:{negative}\:{of}\:{negative} \\ $$$${is}\:{always}\:{positive}….. \\ $$$$\mathcal{B}{ut}\:{I},\:{at}\:{the}\:{moment},\:{not}\:{thinking}\: \\ $$$${mathematically}\:{rigister}\:{it}\: \\ $$$${as}\:{a}\:{complaint}!!! \\ $$
Commented by Tinku Tara last updated on 30/May/20
Hi  Can you update to latest version.  So that flag will not be treated  as like.  If likes>redflags then flag clears.
$$\mathrm{Hi} \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{update}\:\mathrm{to}\:\mathrm{latest}\:\mathrm{version}. \\ $$$$\mathrm{So}\:\mathrm{that}\:\mathrm{flag}\:\mathrm{will}\:\mathrm{not}\:\mathrm{be}\:\mathrm{treated} \\ $$$$\mathrm{as}\:\mathrm{like}. \\ $$$$\mathrm{If}\:\mathrm{likes}>\mathrm{redflags}\:\mathrm{then}\:\mathrm{flag}\:\mathrm{clears}. \\ $$
Commented by Sourav mridha last updated on 30/May/20
wrong!!itshould be (1/(2n))(𝛑/(sin((𝛑/(2n)))))
$$\mathrm{wr}\boldsymbol{{ong}}!!\mathrm{it}\boldsymbol{{should}}\:\boldsymbol{{be}}\:\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\frac{\boldsymbol{\pi}}{\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}}\right)} \\ $$
Commented by mr W last updated on 31/May/20
rasheed sir:   no need to be sorry sir! the app did  what you didn′t want to do. i thought  that guy changed the red flag by  giving himself a like, then i′d like  to know why he did that.  i hope the bug in app is fixed now.
$${rasheed}\:{sir}:\: \\ $$$${no}\:{need}\:{to}\:{be}\:{sorry}\:{sir}!\:{the}\:{app}\:{did} \\ $$$${what}\:{you}\:{didn}'{t}\:{want}\:{to}\:{do}.\:{i}\:{thought} \\ $$$${that}\:{guy}\:{changed}\:{the}\:{red}\:{flag}\:{by} \\ $$$${giving}\:{himself}\:{a}\:{like},\:{then}\:{i}'{d}\:{like} \\ $$$${to}\:{know}\:{why}\:{he}\:{did}\:{that}. \\ $$$${i}\:{hope}\:{the}\:{bug}\:{in}\:{app}\:{is}\:{fixed}\:{now}. \\ $$
Commented by Rasheed.Sindhi last updated on 31/May/20
Thanks Sir! For this reason I said  that a person shoudn′t have power  to like his own post!
$$\mathcal{T}{hanks}\:\mathcal{S}{ir}!\:{For}\:{this}\:{reason}\:{I}\:{said} \\ $$$${that}\:{a}\:{person}\:{shoudn}'{t}\:{have}\:{power} \\ $$$${to}\:{like}\:{his}\:{own}\:{post}! \\ $$
Commented by Tinku Tara last updated on 31/May/20
Please update to latest version from playstore or www.tinkutara.com Version 2.079.
Answered by Sourav mridha last updated on 30/May/20
∫_0 ^∞ (dx/(1+(x^n )^2 ))  replached x^n  by tan𝛂,after that  your integral looks like   (1/n)∫_0 ^(π/2) (sin𝛂)^((1/n)−1) .(cos𝛂)^(1−(1/n)) d𝛂  =(1/(2n))((𝚪((1/(2n))).𝚪(1−(1/(2n))))/(𝚪(1)))=(1/(2n))(𝛑/(sin((𝛑/(2n))))).  remember it is only possible  when     (1/n)=m<1.
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{1}+\left(\mathrm{x}^{\mathrm{n}} \right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{replached}}\:\boldsymbol{{x}}^{\boldsymbol{{n}}} \:\boldsymbol{{by}}\:\boldsymbol{{tan}\alpha},\boldsymbol{{after}}\:\boldsymbol{{that}} \\ $$$$\boldsymbol{{your}}\:\boldsymbol{{integral}}\:\boldsymbol{{looks}}\:\boldsymbol{{like}} \\ $$$$\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\boldsymbol{{sin}\alpha}\right)^{\frac{\mathrm{1}}{\boldsymbol{{n}}}−\mathrm{1}} .\left(\boldsymbol{{cos}\alpha}\right)^{\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{{n}}}} \boldsymbol{{d}\alpha} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\right).\boldsymbol{\Gamma}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}}\frac{\boldsymbol{\pi}}{{sin}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}}\right)}. \\ $$$$\mathrm{re}\boldsymbol{{member}}\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{only}}\:\boldsymbol{{possible}} \\ $$$$\boldsymbol{{when}}\:\:\:\:\:\frac{\mathrm{1}}{\boldsymbol{{n}}}=\boldsymbol{{m}}<\mathrm{1}. \\ $$
Answered by mathmax by abdo last updated on 30/May/20
changement  x^(2n)  =t give x =t^(1/(2n))  ⇒ I_n =(1/(2n))∫_0 ^∞    (t^((1/(2n))−1) /(1+t))dt  =(1/(2n))×(π/(sin((π/(2n)))))⇒ ∫_0 ^∞    (dx/(1+x^(2n) )) =(π/(2nsin((π/(2n)))))  i have used that ∫_0 ^∞   (t^(a−1) /(1+t))dt =(π/(sin(πa)))  if 0<a<1(result proved)
$$\mathrm{changement}\:\:\mathrm{x}^{\mathrm{2n}} \:=\mathrm{t}\:\mathrm{give}\:\mathrm{x}\:=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}} \:\Rightarrow\:\mathrm{I}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2n}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2n}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2n}}×\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)}\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2n}} }\:=\frac{\pi}{\mathrm{2nsin}\left(\frac{\pi}{\mathrm{2n}}\right)} \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{used}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\:\mathrm{if}\:\mathrm{0}<\mathrm{a}<\mathrm{1}\left(\mathrm{result}\:\mathrm{proved}\right) \\ $$

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