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

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 (Q 86484)

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 .

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

I t w a s 29 t h J u n e m a y b e w h e n i f i r s t e n t e r e d . .

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

Y o u d r e w o n J u l y

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

Y o u r Q u e s t i o n i n a n o t h e r f o r m

\int \frac{x^{m}}{1 + x^{2 m}} d x = \frac{1}{m} \int \frac{u^{\frac{1}{m}}}{1 + u^{2}} d u = \frac{1}{2 m} \int \frac{t^{\frac{1}{2 m} - \frac{1}{2}}}{1 + t} d t

= \frac{1}{2 m} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}_{n}}{n!} {(- 1)}^{n} t^{n + \frac{1}{2 m} - \frac{1}{2}} = \frac{t^{\frac{1}{2 m} + \frac{1}{2}}}{2 m} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}_{n} {(- 1)}^{n}}{n! (n + \frac{1}{2 m} + \frac{1}{2})} t^{n}

= \frac{t^{\frac{1}{2 m} + \frac{1}{2}}}{2 m} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}_{n} {(- 1)}^{n} Γ (n + \frac{1}{2 m} + \frac{1}{2})}{n! Γ (n + \frac{1}{2 m} + \frac{3}{2})} t^{n}

= \frac{t^{\frac{1}{2 m} + \frac{1}{2}}}{2 m} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}_{n} {(\frac{1}{2 m} + \frac{1}{2})}_{n} Γ (\frac{1}{2 m} + \frac{1}{2})}{n! (\frac{1}{2 m} + \frac{3}{2}) Γ (\frac{1}{2 m} + \frac{3}{2})} {(- t)}^{n}

= \frac{t^{\frac{m + 1}{2 m}}}{m + 1} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}_{n} {(\frac{m + 1}{2 m})}_{n}}{n! {(\frac{3 m + 1}{2 m})}_{n}} {(- t)}^{n} = \frac{t^{\frac{m + 1}{2 m}}}{m + 1}_{2} F_{1} (- 1, \frac{m + 1}{2 m}; \frac{3 m + 1}{2 m}; - t)

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

H a p p y^{'} b i r t h d a y^{'} t o y o u!

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��