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

Commented by Farruxjano last updated on 30/May/20

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

Commented by john santu last updated on 30/May/20

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{1+x^{2n}}=\frac{\pi i}{2n}.\frac{\left(e^{\frac{\pi i}{2n}}\right)}{(e^{\frac{\pi i}{n}}-1)}$$$$=\frac{\pi i}{2n}.\frac{e^{\frac{\pi i}{2n}}}{e^{\frac{\pi i}{2n}}(e^{\frac{\pi i}{2n}}-e^{-\frac{\pi i}{2n}})}$$$$=\frac{\pi }{2\mathrm{n}}.\frac{i}{e^{\frac{\pi i}{2n}}-e^{-\frac{\pi i}{2n}}}=\frac{\pi }{2\mathrm{n}}.\frac{1}{\sin \left(\frac{\pi }{2\mathrm{n}}\right)}$$$$=\frac{\pi }{2\mathrm{n}.\sin \left(\frac{\pi }{2\mathrm{n}}\right)}.$$

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

$$everyoneknows:whenyouposta$$$$questionhere,youwanttogetanswer.$$$$you{don}^{\prime }tneedtorepeatagainand$$$$againthatyouneedasolution.$$$$$$$$butyoushouldalsoknow:nobodyowes$$$$youanything!youarenotthegod$$$$andallothers{mustn}^{\prime }tdancetoyour$$$$tune!$$$$$$$$besides,itdisturbstheorderofthe$$$$forumwhenyourepeatthesamequestion$$$$byopeningalotofnewthreads.$$

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

$${that}^{\prime }swhyiredflaggedthispostof$$$$you.$$

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

$$isaidihaveredflaggedthispostandi$$$$alsosaidwhy.whoturnedthered$$$$flagto‘‘{like}^{″}againandwhy?$$

Commented by Tinku Tara last updated on 30/May/20

$$\mathrm{Hi}\mathrm{Farruxjano}$$$$\mathrm{I}\mathrm{see}\mathrm{this}\mathrm{is}\mathrm{just}\mathrm{probably}\mathrm{first}\mathrm{post}$$$$\mathrm{from}\mathrm{you}.\mathrm{This}\mathrm{forum}\mathrm{to}\mathrm{enable}$$$$\mathrm{people}\mathrm{to}\mathrm{learn}\mathrm{from}\mathrm{each}\mathrm{other}.$$$$\mathrm{It}\mathrm{is}\mathrm{ok}\mathrm{to}\mathrm{write}\mathrm{part}\mathrm{answers}$$$$\mathrm{to}\mathrm{show}\mathrm{ur}\mathrm{attempt}.$$$$$$$$\mathrm{Just}\mathrm{dont}\mathrm{create}\mathrm{too}\mathrm{many}\mathrm{threads}$$$$\mathrm{for}\mathrm{the}\mathrm{same}\mathrm{question}.$$

Commented by Farruxjano last updated on 30/May/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{terribly}\mathrm{sorry},\mathrm{ok},{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{understood}!$$

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

$$iappreciateitverymuchthatyou$$$$showedyourunderstandingandsaid$$$$sorry.$$

Commented by Rasheed.Sindhi last updated on 30/May/20

$$\mathcal{T}inku\mathcal{T}ara,\mathcal{T}hedeveloper$$$$I,asanexperimentmarkedthe$$$$abovepostasred\mathrm{\_}flagedwhen$$$$itwasalreadyred\mathrm{\_}flaggedby$$$$mrWsir.Theresultwas{}^{\prime }alike$$$$is{added}^{\prime }tothepostandthepost$$$$wasclearofred\mathrm{\_}flag!!!\left(SorrymrW\right)This$$$$behavioroftheappismathematically$$$$correctwherenegativeofnegative$$$$isalwayspositive.....$$$$\mathcal{B}utI,atthemoment,notthinking$$$$mathematicallyrigisterit$$$$asacomplaint!!!$$

Commented by Tinku Tara last updated on 30/May/20

$$\mathrm{Hi}$$$$\mathrm{Can}\mathrm{you}\mathrm{update}\mathrm{to}\mathrm{latest}\mathrm{version}.$$$$\mathrm{So}\mathrm{that}\mathrm{flag}\mathrm{will}\mathrm{not}\mathrm{be}\mathrm{treated}$$$$\mathrm{as}\mathrm{like}.$$$$\mathrm{If}\mathrm{likes}>\mathrm{redflags}\mathrm{then}\mathrm{flag}\mathrm{clears}.$$

Commented by Sourav mridha last updated on 30/May/20

$$\mathrm{wr}\mathit{o}\mathit{n}\mathit{g}!!\mathrm{it}\mathit{s}\mathit{h}\mathit{o}\mathit{u}\mathit{l}\mathit{d}\mathit{b}\mathit{e}\frac{1}{2\mathit{n}}\frac{\mathit{\pi }}{\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{2\mathrm{n}}\right)}$$

Commented by mr W last updated on 31/May/20

$$rasheedsir:$$$$noneedtobesorrysir!theappdid$$$$whatyou{didn}^{\prime }twanttodo.ithought$$$$thatguychangedtheredflagby$$$$givinghimselfalike,then{i}^{\prime }dlike$$$$toknowwhyhedidthat.$$$$ihopethebuginappisfixednow.$$

Commented by Rasheed.Sindhi last updated on 31/May/20

$$\mathcal{T}hanks\mathcal{S}ir!ForthisreasonIsaid$$$$thataperson{shoudn}^{\prime }thavepower$$$$tolikehisownpost!$$

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

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\left({\mathrm{x}}^{\mathrm{n}}\right)}^2}$$$$\mathit{r}\mathit{e}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{h}\mathit{e}\mathit{d}{\mathit{x}}^{\mathit{n}}\mathit{b}\mathit{y}\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha },\mathit{a}\mathit{f}\mathit{t}\mathit{e}\mathit{r}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\mathit{y}\mathit{o}\mathit{u}\mathit{r}\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{g}\mathit{r}\mathit{a}\mathit{l}\mathit{l}\mathit{o}\mathit{o}\mathit{k}\mathit{s}\mathit{l}\mathit{i}\mathit{k}\mathit{e}$$$$\frac{1}{\mathit{n}}{\int }_0^{\frac{\pi }{2}}{\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\right)}^{\frac{1}{\mathit{n}}-1}.{\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\right)}^{1-\frac{1}{\mathit{n}}}\mathit{d}\mathit{\alpha }$$$$=\frac{1}{2\mathit{n}}\frac{\mathbf{\Gamma }\left(\frac{1}{2\mathit{n}}\right).\mathbf{\Gamma }(1-\frac{1}{2\mathit{n}})}{\mathbf{\Gamma }\left(1\right)}=\frac{1}{2\mathit{n}}\frac{\mathit{\pi }}{sin\left(\frac{\mathit{\pi }}{2\mathrm{n}}\right)}.$$$$\mathrm{re}\mathit{m}\mathit{e}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{i}\mathit{t}\mathit{i}\mathit{s}\mathit{o}\mathit{n}\mathit{l}\mathit{y}\mathit{p}\mathit{o}\mathit{s}\mathit{s}\mathit{i}\mathit{b}\mathit{l}\mathit{e}$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\frac{1}{\mathit{n}}=\mathit{m}<1.$$

Answered by mathmax by abdo last updated on 30/May/20

$$\mathrm{changement}{\mathrm{x}}^{2\mathrm{n}}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{t}}^{\frac{1}{2\mathrm{n}}}\Rightarrow {\mathrm{I}}_{\mathrm{n}}=\frac{1}{2\mathrm{n}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{2\mathrm{n}}-1}}{1+\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{2\mathrm{n}}\times \frac{\pi }{\sin \left(\frac{\pi }{2\mathrm{n}}\right)}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2\mathrm{n}}}=\frac{\pi }{2\mathrm{nsin}\left(\frac{\pi }{2\mathrm{n}}\right)}$$$$\mathrm{i}\mathrm{have}\mathrm{used}\mathrm{that}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\mathrm{if}0<\mathrm{a}<1\left(\mathrm{result}\mathrm{proved}\right)$$

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