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On-this-same-day-last-year-I-made-my-first-on-this-forum-Q86484-and-I-was-very-astonished-by-the-answers-which-I-received-I-underestimated-the-place-at-first-before-getting-to-know-it-better-A

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Question Number 137056 by Ar Brandon last updated on 29/Mar/21
On this same day last year I made my first on this forum (Q86484) and I was very astonished by the answers which I received. I underestimated the place at first before getting to know it better. And I realised I knew nothing. I′m grateful with all the teachings which I′ve acquired from you all. You guys are just so amazing. Thank you once more. Special thanks to you too Mr Tinku-Tara. 😉
$$\:\:\:\:\mathrm{On}\:\mathrm{this}\:\mathrm{same}\:\mathrm{day}\:\mathrm{last}\:\mathrm{year}\:\mathrm{I}\:\mathrm{made}\:\mathrm{my}\:\mathrm{first}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:\left(\mathrm{Q86484}\right) \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{was}\:\mathrm{very}\:\mathrm{astonished}\:\mathrm{by}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{which}\:\mathrm{I}\:\mathrm{received}. \\ $$$$\mathrm{I}\:\mathrm{underestimated}\:\mathrm{the}\:\mathrm{place}\:\mathrm{at}\:\mathrm{first}\:\mathrm{before}\:\mathrm{getting}\:\mathrm{to}\:\mathrm{know}\:\mathrm{it} \\ $$$$\mathrm{better}.\:\mathrm{And}\:\mathrm{I}\:\mathrm{realised}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{nothing}.\: \\ $$$$\:\:\:\:\mathrm{I}'\mathrm{m}\:\mathrm{grateful}\:\mathrm{with}\:\mathrm{all}\:\mathrm{the}\:\mathrm{teachings}\:\mathrm{which}\:\mathrm{I}'\mathrm{ve}\:\mathrm{acquired}\:\mathrm{from}\: \\ $$$$\mathrm{you}\:\mathrm{all}.\:\mathrm{You}\:\mathrm{guys}\:\mathrm{are}\:\mathrm{just}\:\mathrm{so}\:\mathrm{amazing}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{once}\:\mathrm{more}. \\ $$$$\mathrm{Special}\:\mathrm{thanks}\:\mathrm{to}\:\mathrm{you}\:\mathrm{too}\:\mathrm{Mr}\:\mathrm{Tinku}-\mathrm{Tara}.\: \\ $$😉
Commented by Dwaipayan Shikari last updated on 29/Mar/21
It was 29th June may be when i first entered ..
$${It}\:{was}\:\mathrm{29}{th}\:{June}\:{may}\:{be}\:{when}\:{i}\:{first}\:{entered}\:.. \\ $$
Commented by Ar Brandon last updated on 29/Mar/21
��You first drew my attention in August.
Commented by Dwaipayan Shikari last updated on 29/Mar/21
You drew on July
$${You}\:{drew}\:{on}\:{July} \\ $$
Commented by Dwaipayan Shikari last updated on 29/Mar/21
Your Question in another form ∫(x^m /(1+x^(2m) ))dx=(1/m)∫(u^(1/m) /(1+u^2 ))du=(1/(2m))∫(t^((1/(2m))−(1/2)) /(1+t))dt =(1/(2m))Σ_(n=0) ^∞ (((−1)_n )/(n!))(−1)^n t^(n+(1/(2m))−(1/2)) =(t^((1/(2m))+(1/2)) /(2m))Σ_(n=0) ^∞ (((−1)_n (−1)^n )/(n!(n+(1/(2m))+(1/2))))t^n =(t^((1/(2m))+(1/2)) /(2m))Σ_(n=0) ^∞ (((−1)_n (−1)^n Γ(n+(1/(2m))+(1/2)))/(n!Γ(n+(1/(2m))+(3/2))))t^n =(t^((1/(2m))+(1/2)) /(2m))Σ_(n=0) ^∞ (((−1)_n ((1/(2m))+(1/2))_n Γ((1/(2m))+(1/2)))/(n! ((1/(2m))+(3/2))Γ((1/(2m))+(3/2))))(−t)^n =(t^((m+1)/(2m)) /(m+1))Σ_(n=0) ^∞ (((−1)_n (((m+1)/(2m)))_n )/(n!(((3m+1)/(2m)))_n ))(−t)^n =(t^((m+1)/(2m)) /(m+1)) _2 F_1 (−1,((m+1)/(2m));((3m+1)/(2m));−t)
$${Your}\:{Question}\:{in}\:{another}\:{form} \\ $$$$\int\frac{{x}^{{m}} }{\mathrm{1}+{x}^{\mathrm{2}{m}} }{dx}=\frac{\mathrm{1}}{{m}}\int\frac{{u}^{\frac{\mathrm{1}}{{m}}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\frac{\mathrm{1}}{\mathrm{2}{m}}\int\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}{m}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{m}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)_{{n}} }{{n}!}\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\frac{\mathrm{1}}{\mathrm{2}{m}}−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{m}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)_{{n}} \left(−\mathrm{1}\right)^{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{t}^{{n}} \\ $$$$=\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{m}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)_{{n}} \left(−\mathrm{1}\right)^{{n}} \Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{{n}!\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{t}^{{n}} \\ $$$$=\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{m}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{{n}!\:\left(\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{m}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\left(−{t}\right)^{{n}} \\ $$$$=\frac{{t}^{\frac{{m}+\mathrm{1}}{\mathrm{2}{m}}} }{{m}+\mathrm{1}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)_{{n}} \left(\frac{{m}+\mathrm{1}}{\mathrm{2}{m}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{3}{m}+\mathrm{1}}{\mathrm{2}{m}}\right)_{{n}} }\left(−{t}\right)^{{n}} =\frac{{t}^{\frac{{m}+\mathrm{1}}{\mathrm{2}{m}}} }{{m}+\mathrm{1}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(−\mathrm{1},\frac{{m}+\mathrm{1}}{\mathrm{2}{m}};\frac{\mathrm{3}{m}+\mathrm{1}}{\mathrm{2}{m}};−{t}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 29/Mar/21
Happy ′ birth day ′ to you!
$$\mathcal{H}{appy}\:'\:{birth}\:{day}\:'\:{to}\:{you}! \\ $$
Commented by Ar Brandon last updated on 29/Mar/21
Hihihi ����
Commented by Ar Brandon last updated on 29/Mar/21
Yeah, birth on forum, haha ! ��Thanks Sir
Commented by Rasheed.Sindhi last updated on 29/Mar/21
��I am trying to find emojies of a weeping boy and feeder ���� (You are one year old!)
Commented by Rasheed.Sindhi last updated on 29/Mar/21
(pl excuse me if you don`t like this joke)
Commented by Ar Brandon last updated on 29/Mar/21
��������
Commented by Rasheed.Sindhi last updated on 29/Mar/21
BTW I apreciate your contribution towards this forum and your sense of maths. (I say same thing about some other youngers like Dwipayan, bemath.)
Commented by Ar Brandon last updated on 29/Mar/21
I too appreciate them. I learned a lot from Shikari, despite the fact that he's a year younger than me��

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