Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Some problem occured in proving the following reduction formula.

$$ \\ I_{(m,n)}\; =\;\int x^m(x+a)^ndx\; = \; \frac{x^m(x+a)^{n+1}}{m+n+1}-\frac{ma}{m+n+1}I_{(m-1,n)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m,n \in N $$ I have tried by using integration by part,here are my result

$$ \begin{align} I_{(m,n)}\; =\;\frac{x^{m+1}(x+a)^n}{m+1}-\frac{n}{m+1}I_{(m+1,n-1)}\\ I_{(m,n)}\; =\;\frac{x^{m}(x+a)^{n+1}}{n+1}-\frac{m}{n+1}I_{(m-1,n+1)} \end{align} $$ I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong. Any help would be appreciated.Thank you.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Hint: $I(m,n)=\int x^m(x+a)^ndx=\int x^{m-1}(x+a-a)(x+a)^{n}dx=\int x^{m-1}(x+a)^{n+1}dx-a\int x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$

Use this with your second reduction formula

share|improve this answer
    
why would you think in that way?Is there any hint motivated us to do that? –  Vulcan Dec 7 '12 at 12:43
    
The reduction fromulae you got will not help you reach your goal, thus its time to try something else –  Amr Dec 7 '12 at 12:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.