A cot integral limit

Evaluate the limit:

Solution

Recalling the Taylor expansion of around we have that

Thus,

The limit follows to be .

Definite integral

Let be a function such that

Evaluate the integral .

Solution

We multiply the given equation by . Thus,

Let and .

We now deal with the first integral:

As for the second integral we have:

Hence,

Offset logarithmic integral inequality

Prove that

Solution

We have successively:

Limit of geometric mean of binomial coefficients

Let denote the geometric mean of the binomial coefficients

Prove that .

Solution

We note that

On the other hand the following lemma holds:

Lemma: Let be a monotonic function. It holds that

Proof: Due to monotony it holds that for . Hence summing over all these values of k we get that

The result follows.

Applying the above to on we get that:

Thus,

(1)

Similarly, applying the above to on we get that:

(2)

The result follows.

Digamma and Trigamma functions

Let and denote the digamma and trigamma functions respectively. Prove that:

where denotes the Euler – Mascheroni constant.

Solution

We begin with the recently discovered identity:

Letting we get that

Now combining this result here we conclude the exercise.

Who is Tolaso?

Find out more at his Encyclopedia Page.