Question-96148

Question Number 96148 by Farruxjano last updated on 30/May/20

Commented by Farruxjano last updated on 30/May/20

$$\mathrm{Please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{solution}\:\::\left(\right. \\ $$

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

$${everyone}\:{knows}:\:{when}\:{you}\:{post}\:{a} \\ $$$${question}\:{here},\:{you}\:{want}\:{to}\:{get}\:{answer}. \\ $$$${you}\:{don}'{t}\:{need}\:{to}\:{repeat}\:{again}\:{and} \\ $$$${again}\:{that}\:{you}\:{need}\:{a}\:{solution}. \\ $$$$ \\ $$$${but}\:{you}\:{should}\:{also}\:{know}:\:{nobody}\:{owes} \\ $$$${you}\:{anything}!\:{you}\:{are}\:{not}\:{the}\:{god} \\ $$$${and}\:{all}\:{others}\:{mustn}'{t}\:{dance}\:{to}\:{your} \\ $$$${tune}! \\ $$$$ \\ $$$${besides},\:{it}\:{disturbs}\:{the}\:{order}\:{of}\:{the} \\ $$$${forum}\:{when}\:{you}\:{repeat}\:{the}\:{same}\:{question} \\ $$$${by}\:{opening}\:{alot}\:{of}\:{new}\:{threads}. \\ $$

Commented by mr W last updated on 30/May/20

$${that}'{s}\:{why}\:{i}\:{red}\:{flagged}\:{this}\:{post}\:{of} \\ $$$${you}. \\ $$

Commented by mr W last updated on 30/May/20

$${i}\:{said}\:{i}\:{have}\:{red}\:{flagged}\:{this}\:{post}\:{and}\:{i} \\ $$$${also}\:{said}\:{why}.\:{who}\:{turned}\:{the}\:{red} \\ $$$${flag}\:{to}\:“{like}''\:{again}\:{and}\:{why}? \\ $$

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

$${i}\:{appreciate}\:{it}\:{very}\:{much}\:{that}\:{you} \\ $$$${showed}\:{your}\:{understanding}\:{and}\:{said} \\ $$$${sorry}. \\ $$

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

$${rasheed}\:{sir}:\: \\ $$$${no}\:{need}\:{to}\:{be}\:{sorry}\:{sir}!\:{the}\:{app}\:{did} \\ $$$${what}\:{you}\:{didn}'{t}\:{want}\:{to}\:{do}.\:{i}\:{thought} \\ $$$${that}\:{guy}\:{changed}\:{the}\:{red}\:{flag}\:{by} \\ $$$${giving}\:{himself}\:{a}\:{like},\:{then}\:{i}'{d}\:{like} \\ $$$${to}\:{know}\:{why}\:{he}\:{did}\:{that}. \\ $$$${i}\:{hope}\:{the}\:{bug}\:{in}\:{app}\:{is}\:{fixed}\:{now}. \\ $$

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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