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Question Number 137056 by Ar Brandon last updated on 29/Mar/21

$$\mathrm{On}\mathrm{this}\mathrm{same}\mathrm{day}\mathrm{last}\mathrm{year}\mathrm{I}\mathrm{made}\mathrm{my}\mathrm{first}\mathrm{on}\mathrm{this}\mathrm{forum}\left(\mathrm{Q}86484\right)$$$$\mathrm{and}\mathrm{I}\mathrm{was}\mathrm{very}\mathrm{astonished}\mathrm{by}\mathrm{the}\mathrm{answers}\mathrm{which}\mathrm{I}\mathrm{received}.$$$$\mathrm{I}\mathrm{underestimated}\mathrm{the}\mathrm{place}\mathrm{at}\mathrm{first}\mathrm{before}\mathrm{getting}\mathrm{to}\mathrm{know}\mathrm{it}$$$$\mathrm{better}.\mathrm{And}\mathrm{I}\mathrm{realised}\mathrm{I}\mathrm{knew}\mathrm{nothing}.$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{grateful}\mathrm{with}\mathrm{all}\mathrm{the}\mathrm{teachings}\mathrm{which}{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{acquired}\mathrm{from}$$$$\mathrm{you}\mathrm{all}.\mathrm{You}\mathrm{guys}\mathrm{are}\mathrm{just}\mathrm{so}\mathrm{amazing}.\mathrm{Thank}\mathrm{you}\mathrm{once}\mathrm{more}.$$$$\mathrm{Special}\mathrm{thanks}\mathrm{to}\mathrm{you}\mathrm{too}\mathrm{Mr}\mathrm{Tinku}-\mathrm{Tara}.$$😉

Commented by Dwaipayan Shikari last updated on 29/Mar/21

$$Itwas29thJunemaybewhenifirstentered..$$

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

$$YoudrewonJuly$$

Commented by Dwaipayan Shikari last updated on 29/Mar/21

$$YourQuestioninanotherform$$$$\int \frac{x^m}{1+x^{2m}}dx=\frac{1}{m}\int \frac{u^{\frac{1}{m}}}{1+u^2}du=\frac{1}{2m}\int \frac{t^{\frac{1}{2m}-\frac{1}{2}}}{1+t}dt$$$$=\frac{1}{2m}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}_n}{n!}{(-1)}^n{t}^{n+\frac{1}{2m}-\frac{1}{2}}=\frac{t^{\frac{1}{2m}+\frac{1}{2}}}{2m}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}_n{(-1)}^n}{n!(n+\frac{1}{2m}+\frac{1}{2})}{t}^n$$$$=\frac{t^{\frac{1}{2m}+\frac{1}{2}}}{2m}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}_n{(-1)}^n\mathrm{\Gamma }(n+\frac{1}{2m}+\frac{1}{2})}{n!\mathrm{\Gamma }(n+\frac{1}{2m}+\frac{3}{2})}{t}^n$$$$=\frac{t^{\frac{1}{2m}+\frac{1}{2}}}{2m}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}_n{(\frac{1}{2m}+\frac{1}{2})}_n\mathrm{\Gamma }(\frac{1}{2m}+\frac{1}{2})}{n!(\frac{1}{2m}+\frac{3}{2})\mathrm{\Gamma }(\frac{1}{2m}+\frac{3}{2})}{(-t)}^n$$$$=\frac{t^{\frac{m+1}{2m}}}{m+1}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}_n{\left(\frac{m+1}{2m}\right)}_n}{n!{\left(\frac{3m+1}{2m}\right)}_n}{(-t)}^n=\frac{t^{\frac{m+1}{2m}}}{m+1}{}_2{F}_1(-1,\frac{m+1}{2m};\frac{3m+1}{2m};-t)$$

Commented by Rasheed.Sindhi last updated on 29/Mar/21

$$\mathcal{H}appy{}^{\prime }birthday{}^{\prime }toyou!$$

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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