I bet there is a better way , but I thought this was fun.
First lets break this up a bit
Lets focus on the 2nd integral to start
so if we integrate by parts with and
If we integrate by parts again we get
Doing this n-1 times we get
Now lets look at the last bit
if we integrate this by parts we get
Now lets integrate the last integral by parts again with
So finally we get
Now subbing this back into the very first line we get
evaluateing the last integral and simplifying gives
Hello,
Here is what I tried...
By differentiating this (under the integral sign), we can get :
But the latter is exactly
And we also have
So finally :
And :
That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute ... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.
I'll think about it later. See ya
here Moo actually used by parts: and the fact that
this is a nice approach. you probably don't need to think about finding because 's solution, which is exactly how i created the problem, has already answered your question.
But the latter is exactly
And we also have
So finally :
And :
That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute ... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.
I'll think about it later. See ya