The integral logarithm is defined for by
Prove the following properties of .
- .
- where is a constant depending on . Find the value of for each .
- Prove there exists a constant such that
and find the value of this constant.
- Let . Find an expression for
in terms of .
- Define a function for by
Prove that
- Proof. We derive this by integrating by parts. Let
Then we have
- Proof. The proof is by induction. Starting with part (a) we have
To evaluate the integral in this expression we integrate by parts with
This gives us
Therefore we have
where . This is the case . Now, assume the formula hold for some integer . Then we have
We then evaluate the integral in this expression using integration by parts, as before, let
Therefore, we have
Plugging this back into the expression we had from the induction hypothesis we obtain
Therefore, the formula holds for the case , and hence, for all integers , where
- Proof. We start with the definition of the integral logarithm,
and make the substitution , . This gives us . Therefore,
where is a constant
- (Note: In the comments, tom correctly suggests an easier way to do this is to use part (c) along with translation and expansion/contraction of the integral. The way I have here works also, but requires an inspired choice of substitution.) We start with the given integral,
and make the substitution
Therefore, using the given fact that , we have
- From part (d) we know that
Then, for the term we consider the integral
where . Similar to part (d) we make the substitution,
This gives us
Therefore, we have
Taking the derivative we then have
Part (d) can be done using the translation and expansion properties along with part (c). Your substitution was interesting, but at this stage might have eluded me.
Though admittedly it would be more difficult to guess the identity in part (c) so your solution is appreciated.
Oh, there was also a typo in part (d). I had something an integral of instead of . Fixed now. Anyway, you’re right doing this using part (c) is easier. I’m not sure why I went with this substitution… it is not an obvious one to make at all. (I’m sure that I chose it because of the value of the constant we are given in the problem.) I’ll put in a note directing people to your comment, but I’ll leave the substitution solution up there since it’s still correct and maybe people will like to see an overly complicated way to do it.