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DreamWeaver
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This is not so much a tutorial, but rather a collection of useful results and techniques. Some of the proofs will be quite long, since as much as possible, I'll aim to prove most results and functional relations used herein, rather than just present another's identity as fact. There will be a few exceptions - to avoid clutter - but not many (and they'll be along more basic lines, like Leibniz's rule for differentiation of a product). -------------------A wide variety of logarithmic integrals can be obtained by evaluating a simpler parametric integral, and then differentiating that w.r.t the parameter in question. Here's the simplest example of all...
Let \(\displaystyle 0 < z \le 1\,\) and \(\displaystyle q > 0\,\) be real numbers, then\(\displaystyle \mathcal{I}(z,q)=\int_0^zx^{q-1}\,dx=\frac{z^q}{q}\)Repeated differentiation of this integral w.r.t. the parameter \(\displaystyle q\,\) then gives:\(\displaystyle \mathcal{I}^{(m)}(z,q)=\frac{d^m}{dq^m}\,\int_0^zx^{q-1}\,dx=\int_0^z\left[\frac{d^m}{dq^m}\,x^{q-1}\right]\,dx=\)\(\displaystyle \int_0^zx^{q-1}(\log x)^m=\frac{d^m}{dq^m}\,\frac{z^q}{q}=\frac{(-1)^mm!\,z^q}{q^{m+1}}\)This simple result can be used to help evaluate vast numbers of considerably more complicated logarithmic integrals. The general idea is the same, however. Let's say we have a continuous function \(\displaystyle \phi(x)\,\), such that the following integral is meaningful (and at worst has singularities near \(\displaystyle x=0\,\), or \(\displaystyle x=z\,\), but nowhere in between):
\(\displaystyle \mathcal{P} (q,z)=\int_0^zx^{q-1}\,\phi(x)\,dx\)If we can evaluate this integral, the differentiation of that result will give us a closed form for integrals of the type\(\displaystyle \mathcal{P}^{(m)}(q,z)=\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx\)and\(\displaystyle \mathcal{P}^{(m)}(1,z)=\lim_{q\to 1}\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx=\int_0^1(\log x)^m\,\phi(x)\,dx\)-------------------
Here's an example. Let \(\displaystyle \phi(x)=\log(1-x^p)\,\), for \(\displaystyle p\in \mathbb{R}^+\,\). Using the series definition for the logarithm
\(\displaystyle \log(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}\)We get
\(\displaystyle \int_0^zx^{q-1}\log(1-x^p)\,dx=-\sum_{k=1}^{\infty}\frac{1}{k}\int_0^zx^{kp+q-1}\,dx=\)\(\displaystyle -\sum_{k=1}^{\infty}\frac{1}{k}\left[\frac{x^{kp+q}}{(kp+q)}\right]_0^z=-\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k(kp+q)}=
-\sum_{k=1}^{\infty}\frac{\frac{1}{q}[(kp+q)-kp]}{k(kp+q)}z^{kp+q}=\)\(\displaystyle -\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{p}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(kp+q)}=
-\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(k+q/p)}\)Now, if we let \(\displaystyle z=1\,\), then those last two series are expressible in terms of the polygamma function. The polygamma function has the series representation\(\displaystyle \psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\)Or, rearranging the terms\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)=\gamma+\psi_0(z)\)Where \(\displaystyle \gamma\,\) is the Euler-Mascheroni constant. Using this definition, we can re-write our series - in the case \(\displaystyle z=1\,\) - as:\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\,\left[\sum_{k=1}^{\infty}\frac{1}{k}- \sum_{k=1}^{\infty}\frac{1}{(k+q/p)}\right]=\)\(\displaystyle -\frac{1}{q}\,\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1+\left(\frac{p+q}{p}\right)}\right)=\)\(\displaystyle -\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)-------------------Before we move on, a quick word about the polygamma function. By definition, the polygamma function is the logarithmic derivative of the gamma function:\(\displaystyle \psi_0(z)=\frac{d}{dz}\, \log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}\)Similarly, higher order polygamma functions are defined as the higher order derivatives of the loggamma function \(\displaystyle \log\Gamma(x)\,\), so for \(\displaystyle m\ge 1\,\)\(\displaystyle \psi_m(z)=\frac{d^{m+1}}{dz^{m+1}}\log\Gamma(z)= \frac{d^m}{dz^m} \psi_0(z)=\)\(\displaystyle \frac{d^m}{dz^m} \left[\psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\right]=\)\(\displaystyle \frac{d^m}{dz^m}\sum_{n=1}^{\infty}\frac{1}{z+n-1}=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)We now have a series expansion for the higher order polygamma functions:\(\displaystyle \psi_{m \ge 1}(z)=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)-------------------
Returning to our logarithmic integral\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)We can now differentiate both sides of this equation m-times with respect to q, to obtain\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=-\frac{d^m}{dq^m}\,\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)To evaluate the RHS, we use Leibniz's rule for differentiation of a product:\(\displaystyle \frac{d^m}{dx^m} [f(x)\, g(x)]=\sum_{j=0}^m\binom{m}{j}\left[\frac{d^j}{dx^j}\, f(x)\right]\, \left[\frac{d^{m-j}}{dx^{m-j}}\, g(x)\right]\)
\(\displaystyle \Rightarrow\)\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)\(\displaystyle \left(-\gamma\, \frac{d^m}{dq^m}\,\frac{1}{q}\right)-\sum_{j=0}^m\binom{m}{j} \left(\frac{d^{m-j}}{dx^{m-j}}\,\frac{1}{q}\right)\left[\frac{d^j}{dx^j}\, \psi_0 {\left(\frac{p+q}{p}\right)} \right]=\)\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\binom{m}{j}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\frac{m!}{j!\,(m-j)!}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)
-------------------Final Result:
\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)
\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)
Comments and/or questions should be posted here:
http://mathhelpboards.com/commentary-threads-53/commentary-logarithmic-integrals-polylogarithms-associated-functions-6524.html
Let \(\displaystyle 0 < z \le 1\,\) and \(\displaystyle q > 0\,\) be real numbers, then\(\displaystyle \mathcal{I}(z,q)=\int_0^zx^{q-1}\,dx=\frac{z^q}{q}\)Repeated differentiation of this integral w.r.t. the parameter \(\displaystyle q\,\) then gives:\(\displaystyle \mathcal{I}^{(m)}(z,q)=\frac{d^m}{dq^m}\,\int_0^zx^{q-1}\,dx=\int_0^z\left[\frac{d^m}{dq^m}\,x^{q-1}\right]\,dx=\)\(\displaystyle \int_0^zx^{q-1}(\log x)^m=\frac{d^m}{dq^m}\,\frac{z^q}{q}=\frac{(-1)^mm!\,z^q}{q^{m+1}}\)This simple result can be used to help evaluate vast numbers of considerably more complicated logarithmic integrals. The general idea is the same, however. Let's say we have a continuous function \(\displaystyle \phi(x)\,\), such that the following integral is meaningful (and at worst has singularities near \(\displaystyle x=0\,\), or \(\displaystyle x=z\,\), but nowhere in between):
\(\displaystyle \mathcal{P} (q,z)=\int_0^zx^{q-1}\,\phi(x)\,dx\)If we can evaluate this integral, the differentiation of that result will give us a closed form for integrals of the type\(\displaystyle \mathcal{P}^{(m)}(q,z)=\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx\)and\(\displaystyle \mathcal{P}^{(m)}(1,z)=\lim_{q\to 1}\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx=\int_0^1(\log x)^m\,\phi(x)\,dx\)-------------------
Here's an example. Let \(\displaystyle \phi(x)=\log(1-x^p)\,\), for \(\displaystyle p\in \mathbb{R}^+\,\). Using the series definition for the logarithm
\(\displaystyle \log(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}\)We get
\(\displaystyle \int_0^zx^{q-1}\log(1-x^p)\,dx=-\sum_{k=1}^{\infty}\frac{1}{k}\int_0^zx^{kp+q-1}\,dx=\)\(\displaystyle -\sum_{k=1}^{\infty}\frac{1}{k}\left[\frac{x^{kp+q}}{(kp+q)}\right]_0^z=-\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k(kp+q)}=
-\sum_{k=1}^{\infty}\frac{\frac{1}{q}[(kp+q)-kp]}{k(kp+q)}z^{kp+q}=\)\(\displaystyle -\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{p}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(kp+q)}=
-\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(k+q/p)}\)Now, if we let \(\displaystyle z=1\,\), then those last two series are expressible in terms of the polygamma function. The polygamma function has the series representation\(\displaystyle \psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\)Or, rearranging the terms\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)=\gamma+\psi_0(z)\)Where \(\displaystyle \gamma\,\) is the Euler-Mascheroni constant. Using this definition, we can re-write our series - in the case \(\displaystyle z=1\,\) - as:\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\,\left[\sum_{k=1}^{\infty}\frac{1}{k}- \sum_{k=1}^{\infty}\frac{1}{(k+q/p)}\right]=\)\(\displaystyle -\frac{1}{q}\,\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1+\left(\frac{p+q}{p}\right)}\right)=\)\(\displaystyle -\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)-------------------Before we move on, a quick word about the polygamma function. By definition, the polygamma function is the logarithmic derivative of the gamma function:\(\displaystyle \psi_0(z)=\frac{d}{dz}\, \log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}\)Similarly, higher order polygamma functions are defined as the higher order derivatives of the loggamma function \(\displaystyle \log\Gamma(x)\,\), so for \(\displaystyle m\ge 1\,\)\(\displaystyle \psi_m(z)=\frac{d^{m+1}}{dz^{m+1}}\log\Gamma(z)= \frac{d^m}{dz^m} \psi_0(z)=\)\(\displaystyle \frac{d^m}{dz^m} \left[\psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\right]=\)\(\displaystyle \frac{d^m}{dz^m}\sum_{n=1}^{\infty}\frac{1}{z+n-1}=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)We now have a series expansion for the higher order polygamma functions:\(\displaystyle \psi_{m \ge 1}(z)=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)-------------------
Returning to our logarithmic integral\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)We can now differentiate both sides of this equation m-times with respect to q, to obtain\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=-\frac{d^m}{dq^m}\,\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)To evaluate the RHS, we use Leibniz's rule for differentiation of a product:\(\displaystyle \frac{d^m}{dx^m} [f(x)\, g(x)]=\sum_{j=0}^m\binom{m}{j}\left[\frac{d^j}{dx^j}\, f(x)\right]\, \left[\frac{d^{m-j}}{dx^{m-j}}\, g(x)\right]\)
\(\displaystyle \Rightarrow\)\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)\(\displaystyle \left(-\gamma\, \frac{d^m}{dq^m}\,\frac{1}{q}\right)-\sum_{j=0}^m\binom{m}{j} \left(\frac{d^{m-j}}{dx^{m-j}}\,\frac{1}{q}\right)\left[\frac{d^j}{dx^j}\, \psi_0 {\left(\frac{p+q}{p}\right)} \right]=\)\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\binom{m}{j}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\frac{m!}{j!\,(m-j)!}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)
-------------------Final Result:
\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)
\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)
Comments and/or questions should be posted here:
http://mathhelpboards.com/commentary-threads-53/commentary-logarithmic-integrals-polylogarithms-associated-functions-6524.html
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