Question

show that \[\large \int\limits_0^{\infty} \dfrac{x^{a}\,dx}{1+x^b} =
\frac{\pi}{b\sin(\pi(a+1)/b )}\]

\(a\ge 0\) and \(b\gt a+1\)

Answer 1

what have u played with beta :O

Answer 2

kindof.. its going to be simple with a clever substitution

Answer 3

btw that integral only converges for b>a+1 by comparison test so assume thats the case

Answer 4

\((1+x)^b=1+x^b+\sum _{k=1}^{b-1}x^{k}\\
1+x^b=\sum _{k=1}^{b-1}x^{k}-(1+x)^b \)

Answer 5

Interesting, but what good is that

Answer 6

\(\large \large \int\limits_0^{\infty} \dfrac{x^{a}\,dx}{1+x^b} =
\large \large \int\limits_0^{\infty} \dfrac{x^{a}\,dx}{\sum _{k=1}^{b-1}x^{k}-(1+x)^b } \)
by partial fraction
\(\large \large \dfrac{x^{a}}{1+x^b} =
\large \large \dfrac{A }{\sum _{k=1}^{b-1}x^{k} } - \dfrac{B }{(1+x)^b } \)

i dont even think if this is possible in this way but i'll give a try

Answer 7

You're assuming a and b are integers

Answer 8

eh , never mind then :\

Answer 9

its okay lets assume they are integers

Answer 10

i dont know how to do the fraction thingy
so i'll try something else

Answer 11

i got something nice

Answer 12

okay im waiting :)

Answer 13

ugh nvm i made mistake somewhere
leave it

Answer 14

Ohk.. I'll leave the question open and check in the morning, cya!

Answer 15

just say we don't need any of those Beta and Gamma functions here

Answer 16

\(\large \text{assume :-} \\ u=x^b , \text{ then } x=u^{\frac{1}{b}}\text{ and } \large \,dx=\frac{1}{b}\frac{du}{x^{b-1}}\\
\\ \large \begin{align*}
\int _0^\infty \frac{x^a}{1+x^b} &= \frac{1}{b} \int _0^\infty \frac{x^a}{(1+u)x^{b-1}}\,du\\
&=\frac{1}{b} \int _0^\infty \frac{x^{a-b+1}}{(1+u) }\,du \\
&= \frac{1}{b} \int _0^\infty \frac{u^{\frac{a +1}{b}-1}}{(1+u) }\,du
\end{align*} \\ \large \text{ let } m =\frac{a+1}{b} \text{ and } n+m=1 \text{ then }n=1- \frac{a+1}{b} \\\large \)

Answer 17

\(\begin{align*}
\frac{1}{b} \int _0^\infty \frac{u^{\frac{a +1}{b}-1}}{(1+u) }\,du &= \frac{1}{b} \int _0^\infty \frac{u^{m-1} }{ (1+u)^{n+m } }\,du \\
&=\frac{1}{b} \beta(m,n) \\
&=\frac{1}{b} \beta(\frac{a+1}{b},1-\frac{a+1}{b}) \\
&= \frac{1}{b} \frac{\Gamma(\frac{a+1}{b})\Gamma( 1-\frac{a+1}{b})}{\Gamma (\frac{a+1}{b}+1-\frac{a+1}{b} )}\\
&=\frac{1}{b} \frac{\Gamma(\frac{a+1}{b})\Gamma( 1-\frac{a+1}{b})}{\Gamma (1 )}\\

&=\frac{1}{b} \frac{\pi}{\sin(\pi (\frac{a+1}{b}))}
\end{align*} \)

Answer 18

yaaaaaaaaaaaaaaaay :3
done

Answer 19

note that
\(\Gamma(1)=1\)

Answer 20

and used Euler's reflection formula for last line

Answer 21

we have to go in the complex plane,namely I set:
\[x ^{(b/2)}=z\]
so your integral becomes:
\[\frac{ 2 }{ b } \int\limits_{0}^{\infty}\frac{ z ^{2a/b+2/b-1} }{ 1+z ^{2} }dz\]
now that integral is equal to:
\[\frac{ 2\pi }{ 2 }(Resf(i)+Resf(-i))\]
where Res is the residual value of f(z) and f(z) is our integrand above.
after a standard calculation, we get:
\[\frac{ \pi b }{ 4 }\left( e ^{i \pi(a+1)/b}-e ^{-i \pi (a+1)/b} \right)=\]
\[=\frac{ \pi b }{4 }*2i \sin \left( \pi \frac{ a+1 }{ b } \right)\]
and taking the real part we can write:
\[\frac{ \pi b }{ 2 }*\sin \left( \pi \frac{ a+1 }{ b } \right)\]

Answer 22

please, note that I got sin at numerator, not at denominator

Answer 23

is it possible that?

Answer 24

wait u have little problem with residues , did u use Taylor series to get them ?

Answer 25

no, only the theory of integration of complex functions

Answer 26

oops.. sorry the coefficient is:
\[\frac{ \pi }{ b }\]
not
\[\frac{ \pi b }{ 2 }\]

Answer 27

Hahahaa @Marki gave u both beta and gamma functions @xapproachesinfinity :P

Answer 28

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