Find .

First, we note via symmetry that

.

Now, recall that it was shown here, that

.

Thus

.

And so we find that

.

Now, we can use our previous results for the partial derivatives of the beta function.

We will need

,

,

,

and .

We have

,

and so

.

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Tags: Beta Function, Digamma Function, Gamma Function, integrals, Logarithms, Math, Monday Math, Polygamma Function

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