# Calculating logarithmic integrals without using the derivatives of Beta function.

How to prove the following generalizations without using the derivatives of beta function:

$$i)\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=(-1)^{m-1}m!\sum_{k=1}^\infty\frac{H_k}{(k+n+1)^{m+1}}\\=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right),\quad n>-2,\quad m\in\mathbb{N}$$

$$ii)\int_0^1\frac{x^n\ln^m(x)\ln^2(1-x)}{1-x}\ dx=(-1)^mm!\sum_{k=1}^\infty\frac{H_k^2-H_k^{(2)}}{(k+n+1)^{m+1}}\\=-\frac1{3}\frac{\partial^m}{\partial n^m}\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right),\quad n>-2,\quad m\in\mathbb{N}$$

The common problem with the derivative of Beta function $$\text{B}(a,b)$$ is the case when $$a$$ or $$b$$ approaches zero because we know that Beta derivative involves $$\psi(a),\psi_1(a),\psi_2(a) ..$$ and $$\psi(b),\psi_1(b),\psi_2(b)...$$ and the limit of these polygamma is undefined when $$a$$ or $$b$$ approaches zero and we will need the help of Wolfram or Mathematica to calculate such derivatives.

Using the above identities will help us avoid this issue as we just need to take the derivative of the harmonic number $$\frac{\partial}{\partial n}H_n^{(a)}=a(\zeta(a+1)-H_n^{(a+1)})$$ or you can simply convert the harmonic number to polygamma function $$\psi_a(n+1)=(-1)^{a+1}a!(\zeta(a+1)-H_n^{(a+1)})$$ as the derivative of polygamma is more straightforward.

I will provide the proofs soon but variant approaches are always appreciated.

• Just a nitpick, the parameter $m$ needs to be greater than zero for the integrals to converge. Oct 22, 2019 at 3:36
• thank you .. I didnt test this case. Oct 22, 2019 at 5:58

The first identity:

From $$\sum_{k=1}^\infty H_kx^k=-\frac{\ln(1-x)}{1-x}$$

It follows that

$$\small{\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=-\sum_{k=1}^\infty H_k\int_0^1x^{n+k}\ln^m(x)\ dx=-(-1)^m m!\sum_{k=1}^\infty\frac{H_k}{(n+k+1)^{m+1}}}\tag1$$

By the master theorem we have

$$\frac{H_n^2+H_n^{(2)}}{2n}=\sum_{k=1}^\infty\frac{H_k}{(k+1)(k+n+1)}\tag2$$

multiply both sides of $$(2)$$ by $$n$$ then differentiate with respect to $$n$$ we get

$$\sum_{k=1}^\infty\frac{H_k}{(n+k+1)^2}=\frac12\frac{\partial}{\partial n}\left(H_n^2+H_n^{(2)}\right)\tag3$$

Now if we differentiate both sides of $$(3)$$ with respect to $$n$$ $$m$$ times we get

$$-(-1)^m m! \sum_{k=1}^\infty\frac{H_k}{(n+k+1)^{m+1}}=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right)\tag4$$ Plug $$(4)$$ in $$(1)$$ we get the first identity.

The second identity:

From the identity

$$\frac{\ln^2(1-x)}{1-x}=\sum_{k=1}^\infty (H_k^2-H_k^{(2)})x^k$$

it follows that

$$\int_0^1\frac{x^n\ln^m(x)\ln^2(1-x)}{1-x}\ dx=(-1)^mm!\sum_{k=1}^\infty\frac{H_k^2-H_k^{(2)}}{(k+n+1)^{m+1}}\tag1$$

Again, by the master theorem we have

$$\sum_{k=1}^\infty \frac{H_k^2-H_k^{(2)}}{(k+1)(k+n+1)}=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{3n}\tag2$$

Multiply both sides of $$(2)$$ by $$n$$ then differentiate with respect to $$n$$ to have

$$\sum_{k=1}^\infty \frac{H_k^2-H_k^{(2)}}{(k+n+1)^2}=\frac13\frac{\partial}{\partial n}\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right)\tag3$$

Now differentiate both sides of $$(3)$$ with respect to $$n$$ $$m$$ times we get

$$-(-1)^mm!\sum_{k=1}^\infty \frac{H_k^2-H_k^{(2)}}{(k+n+1)^{m+1}}=\frac13\frac{\partial^m}{\partial n^m}\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right)\tag4$$

Plug $$(4)$$ in $$(1)$$ we get the second identity

Full credit goes to Cornel as I just generalized his solution to get the second identity. As for proving the first identity, I just followed the same approach.

BONUS:

Similarly, from the identity

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n\tag1$$

It follows that

$$\int_0^1\frac{x^n\ln^m(x)\ln^3(1-x)}{1-x}\ dx=-(-1)^mm!\sum_{k=1}^\infty\frac{H_k^3-3H_k H_k^{(2)}+2H_k^{(3)}}{(k+n+1)^{m+1}}\tag2$$

By the master theorem, we have

$$\small{\sum_{k=1}^\infty\frac{H_k^3-3H_k H_k^{(2)}+2H_k^{(3)}}{(n+1)(k+n+1)}=\frac1{4n}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)}\tag3$$

Multiply both sides of $$(3)$$ by $$n$$ then differentiate with respect to $$n$$ we get

$$\small{\sum_{k=1}^\infty\frac{H_k^3-3H_k H_k^{(2)}+2H_k^{(3)}}{(k+n+1)^2}=\frac1{4}\frac{\partial}{\partial n}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)}\tag4$$

Differentiate both sides of $$(4)$$ with respect to $$n$$ $$m$$ times we get

$$-(-1)^mm!\sum_{k=1}^\infty\frac{H_k^3-3H_k H_k^{(2)}+2H_k^{(3)}}{(k+n+1)^{m+1}}\\=\frac1{4}\frac{\partial^m}{\partial n^m}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)\tag5$$

Plug $$(5)$$ in $$(2)$$ we get

$$\int_0^1\frac{x^n\ln^m(x)\ln^3(1-x)}{1-x}\ dx=-(-1)^mm!\sum_{k=1}^\infty\frac{H_k^3-3H_k H_k^{(2)}+2H_k^{(3)}}{(k+n+1)^{m+1}}\\=\frac1{4}\frac{\partial^m}{\partial n^m}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)$$