Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 4

Year : 2015

Put $bx=\dfrac {at}{1-t}$ when $x=0,t=0,$ when $x=\infty,t=1$

$\therefore bdx=\dfrac {-at\times \dfrac d{dt}(1-t)+a(1-t)\dfrac {dt}{dt}}{(1-t)^2}$

$bdx=\Bigg[\dfrac {tat+a-at}{(a-t)^2}\Bigg] dt$

$bdx=\dfrac a{(1-t)^2}dt$

And

$a+bx=a+\dfrac {at}{1-t}=\dfrac a{1-t}$

$\therefore I=\int\limits_0^1\Bigg(\dfrac ab\Bigg)^{m-1}.\dfrac {t^{m-1}}{(1-t)^{m-1}}\dfrac 1{a^{m+n}}(1-t)^{m+n}\times\dfrac 1b a\dfrac {dt}{1-t^2}$

$=\dfrac 1{a^nb^m}\int\limits_0^1t^{m-1}(1-t)^{n-1} dt$

$=\dfrac 1{a^nb^m}\beta (m,n)$