An Integral from Irresistible Integrals

I have been trying to work out a question from Irresistible Integrals, but I don't know how to proceed. The exact problem is Exercise 13.7.1. It asks us to show that $$\zeta(3)=4-\zeta(2)+\frac{8}{3}\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt$$

The $\zeta$ in question is the Riemann Zeta Function. The book does this integral through something that its calls its "Master Formula," but asks the reader to check it directly, which I assume means to compute the integral directly. The book gives the hint of expanding the integrand as a series, using $$\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}{n}x^n$$

After applying this, I have $$\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt=\sum_{n=0}^\infty\binom{2n}{n}\int_0^{1/2}t^{2n+1}\ln^3(t)dt$$

$$=-\sum_{n=0}^\infty\frac{1}{2^{2n+2}}\binom{2n}{n}\left(\frac{\ln^3(2)}{2n+2}+\frac{3\ln^2(2)}{(2n+2)^2}+\frac{6\ln(2)}{(2n+2)^3}+\frac{6}{(2n+2)^4}\right)$$

(If someone wants to check that to makes sure I'm right, then by all means, go for it) I get stuck from here. I have a few tricks to evaluate the sum for the first two terms, but not the last two terms. If you have any ideas on how to evaluate this sum (or an alternate way of evaluating this integral in the first place), let me know.

Thanks!

Edit: Considering that this integral is being multiplied by $8/3$ in the original problem, it may be convenient to fact or out the $8/3$ to get

$$\frac{8}{3}\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt=$$

$$-\sum_{n=0}^\infty\frac{1}{3\cdot2^{2n-1}}\binom{2n}{n}\left(\frac{\ln^3(2)}{2n+2}+\frac{3\ln^2(2)}{(2n+2)^2}+\frac{6\ln(2)}{(2n+2)^3}+\frac{6}{(2n+2)^4}\right)$$

It also might be more convenient to factor the $2$'s from the denominators and get

$$\frac{8}{3}\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt=$$

$$-\sum_{n=0}^\infty\frac{1}{3\cdot2^{2n}}\binom{2n}{n}\left(\frac{\ln^3(2)}{n+1}+\frac{3}{2}\frac{\ln^2(2)}{(n+1)^2}+\frac{3}{2}\frac{\ln(2)}{(n+1)^3}+\frac{3}{4}\frac{1}{(n+1)^4}\right)$$

It looks a bit nicer, but I still don't know how to proceed.

Edit 2: A good suggestion is to turn this integral into a form of the Beta function. This can be done a few different ways, but if we take $u=4t^2$ we get

$$\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt=\frac{1}{8}\int_0^1(1-u)^{-1/2}\ln^3\left(\sqrt{\frac{u}{4}}\right)du$$

$$=\frac{1}{64}\int_0^1(1-u)^{-1/2}\left(\ln^3(u)-3\ln^2(u)\ln(4)+3\ln(u)\ln^24)-\ln^3(4)\right)du$$

$$=\frac{1}{64}\left(\frac{\partial^3B}{\partial x^3}(1,1/2)-3\ln(4)\frac{\partial^2B}{\partial x^2}(1,1/2)+3\ln^2(4)\frac{\partial B}{\partial x}(1,1/2)-\ln^3(4)B(1,1/2)\right)$$

At this point, derivatives of the Beta function are tricky, but they boil down to derivatives of the $\psi$ function, which I am not very strong in. If someone has a solution from here, that would be handy as well, but I am hoping that someone can give a solution somehow using a series approach, as suggested in the text.

Solution 1:

This integral is quite nice and clearly deserves an answer, I will try my best, but as you will see I have many gaps too, so please feel free to help me improve the answer.

Let us start by denoting the integral as $I$ and substitute $\,\displaystyle{t=\frac{\sin x}{2}\Rightarrow \frac{dt}{\sqrt{1-4t^2}}=\frac12 dx}\,$ which gives: $$\color{purple}{I=\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt=\frac12 \int_0^\frac{\pi}{2}\frac{\sin x}{2}\ln^3\left(\frac{\sin x}{2}\right)dx}$$ $$=\frac14{\color{red}{\int_0^\frac{\pi}{2}\sin x \ln^3 (\sin x)dx}}-\frac{3\ln 2}{4}{\color{blue}{\int_0^\frac{\pi}{2} \sin x\ln^2(\sin x)dx}}$$ $$+\frac{3\ln^2 2}{4}{\color{#085}{\int_0^\frac{\pi}{2}\sin x\ln (\sin x)dx}}-\frac{\ln^3 2}{4}\color{#A00}{\int_0^\frac{\pi}{2}\sin x \,dx} $$ I will use the following definition of Beta$\rightarrow$ Gamma function which follows for $(14)$ from here, namely: $$\bbox[12pt,#000,border:2px solid purple]{\color{orange}{\int_0^\frac{\pi}{2} \sin^{n }x\,dx=\frac12 B\left(\frac{n+1}{2},\frac12\right)=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}}}$$ Although we can ommit using beta function for the easier one's I will just use it for everyone.

Also I will mostly use known properties/relations, such as:$$\Gamma\left(\frac12\right)=\sqrt{\pi},\quad \Gamma(1+x)=x\Gamma(x)=x!$$ And let's start by evaluating the easier one's first: $$\require{cancel}\color{#A00}{I_4=\int_0^\frac{\pi}{2}\sin x \,dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(1\right)}{\Gamma\left(1+\frac12 \right)}=\frac{\cancel{\sqrt{\pi}}}{2}\frac{1}{\frac12\cancel{\Gamma\left(\frac12\right)}}=1}$$ Now when log-sine terms appears we just have to take a derivative with respect to $n$ then set $n=1$.

That follows since $\displaystyle{\frac{\partial}{\partial n}\sin^n x=\sin^n x\ln(\sin x)}$ also for later, we have: $\displaystyle{\frac{\partial^a}{\partial n^a}\sin^n x=\sin^n x\ln^a(\sin x)}$ $$\color{#085}{I_3=\int_0^\frac{\pi}{2}\sin x\ln (\sin x)dx=\frac{\sqrt{\pi}}{2}\frac{\partial }{\partial n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\bigg|_{n=1}}$$ I saw you got to this point looking at your edit. Just a small observation: Don't plugg in $x=1$ before taking the derivatives.

The things that one needs to know in order to take the derivatives is basic logarithm properties such as: $$f=e^{ln f},\quad \ln \left(\frac{p}{q}\right)=\ln p -\ln q,\quad \frac{d}{du} e^{\ln u}=u \frac{d}{du}(\ln u)$$ And some properties about digamma function, mainly: $\displaystyle{\frac{d }{du(x)}\ln\left(\Gamma(u(x))\right)=\psi (u(x))\cdot u'(x)}$.

Let's go back to our problem and use the properties of the logarithms, we have directly: $$\frac{\partial }{\partial n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\frac{\partial }{\partial n} \left(\ln\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\right)$$$$=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\left(\frac{\partial }{\partial n} \ln\Gamma\left(\frac{n+1}{2}\right)-\frac{\partial }{\partial n} \ln\Gamma\left(1+\frac{n}{2}\right)\right)$$ $$=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\left(\frac12\psi\left(\frac{n+1}{2}\right)-\frac12\psi\left(1+\frac{n}{2}\right)\right)$$ $$\boxed{\Rightarrow\frac{\partial }{\partial n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}= \frac12 \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(1+\frac{n}{2}\right)\right)}\tag1$$ Also we will use digamma's reccurence formula now for the second digamma term, namely: $\displaystyle{\psi\left(1+z\right)=\psi(z)+\frac{1}{z}}$. Note that we could have written from the start: $\,\displaystyle{\Gamma\left(1+\frac{n}{2}\right)=\frac{n}{2}\Gamma\left(\frac{n}{2}\right)},\,$ but then taking derivatives would be harder. $$\color{#085}{I_3=\int_0^\frac{\pi}{2}\sin x\ln(\sin x)dx=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{n+1}{2}\right)}{n\Gamma\left(\frac{n}{2}\right)}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac{2}{n}\right)\bigg|_{n=1}}$$ Also since wikipedia is kind we will use those special values directly. $$\Rightarrow\color{#085}{I_3=\frac{\cancel {\sqrt{\pi}}}{2} \frac{\Gamma\left(1\right)}{\cancel{\Gamma\left(\frac{1}{2}\right)}}\left(\psi\left(1\right)-\psi\left(\frac{1}{2}\right)-2\right)=\frac12\left(\cancel{-\gamma} +2\ln 2 +\cancel{\gamma}-2\right)=\ln2-1}$$ Next up is the blue one: $$\color{blue}{I_2=\int_0^\frac{\pi}{2}\sin x\ln^2(\sin x)dx=\frac{\sqrt{\pi}}{2}\frac{\partial^2 }{\partial n^2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\bigg|_{n=1}}$$ Notice that in $(1)$ we already took a derivative so it's enough to take $1$ more directly from there.

First let's notice that we have to take a derivative of the form $(fg)'=f'g+fg'$. Our $f$ here will be the fraction that contains the gamma function and $g$ the one with digamma, for $f'$ we will have a great time since that is exactly the relation in $(1)$ and for $g$.. well after digamma comes trigamma. $$\Rightarrow\frac{\partial^2 }{\partial n^2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}= \frac{\partial }{\partial n}\frac12 {\color{magenta}{\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}}}{\color{#A80}{\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(1+\frac{n}{2}\right)\right)}}$$ $$=\color{magenta}{\frac14\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(1+\frac{n}{2}\right)\right)}\color{#A80}{\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(1+\frac{n}{2}\right)\right)}$$ $$+\color{magenta}{\frac12\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}}\color{#A80}{{\left(\frac12\psi_1\left(\frac{n+1}{2}\right)-\frac12\psi_1\left(1+\frac{n}{2}\right)\right)}}$$ $$=\boxed{\frac14\underbrace{\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}}_{A}\left(\left(\underbrace{\psi\left(\frac{n+1}{2}\right)-\psi\left(1+\frac{n}{2}\right)}_{B}\right)^2+\left(\underbrace{\psi_1\left(\frac{n+1}{2}\right)-\psi_1\left(1+\frac{n}{2}\right)}_{C}\right)\right)}\tag2$$ Now to find $I_3$ we just need to plugg in $n=1$ and use some special values from here. $$\color{blue}{I_3=\cancel{\frac{\sqrt{\pi}}{2}}\cdot \frac14\cdot \cancel{\frac{\Gamma(1)}{\frac12\Gamma\left(\frac12\right)}}\left(\left(2\ln 2-2\right)^2+\left(\zeta(2) - 3 \zeta(2)+4\right)\right)}$$ $$\Rightarrow \color{blue}{\int_0^\frac{\pi}{2} \sin x\ln^2(\sin x)dx=\ln^2 2 -2\ln 2+2 -\frac{\zeta(2)}{2}}$$ Now for the last integral, we need to take a derivative of $(2)$. I will use some shorter notations. $$\boxed{\Rightarrow\frac{\partial^3 }{\partial n^3}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}=\frac14 \cdot\frac12 A B(B^2+C)+\frac14 A\left(2B\cdot \frac12 C +\frac12 D\right)=\frac18A\left(B^3+3BC+D\right)}$$ Where $\displaystyle{D=\psi_2\left(\frac{n+1}{2}\right)-\psi_2\left(1+\frac{n}{2}\right)}$ and $\displaystyle{\psi_2(z)}$ denotes the polygamma function of order $2$.

We also have the recurrence relation: $\displaystyle{\psi_2(1+z)=\psi_2(z)+\frac{2}{z^3}}$. And using those special values, namely: $$A=\frac{2}{\sqrt{\pi}},\quad B=2\ln2-2$$ $$C=4-2\zeta(2),\quad D=12\zeta(3)-16$$ $$\Rightarrow \color{red}{ I_1=\int_0^\frac{\pi}{2} \sin x\ln^3(\sin x)dx=\frac32 \zeta(3)+\frac32 \zeta(2) -\frac32 \zeta(2)\ln 2 -6+\ln^3 2-3\ln^2 2 +6\ln 2}$$ $$\Rightarrow \color{purple}{4I=\int_0^\frac{\pi}{2} \sin x\ln^3\left(\frac{\sin x}{2}\right)dx}=\color{red}{\frac32 \zeta(3)+\frac32 \zeta(2) -\frac32 \zeta(2)\ln 2 -6+\ln^3 2-3\ln^2 2 +6\ln 2}$$ $$-\color{blue}{3\ln^32 +6\ln^2 2 -6\ln 2 +\frac32 \zeta(2)\ln 2}+\color{#085}{3\ln^3 2-3\ln^2 2}-\color{#A00}{\ln^3 2}=\color{black}{\frac32\zeta(3)+\frac32\zeta(2)-6}$$ $$\Rightarrow \boxed{\color{purple}{I=\int_0^{1/2}\frac{t\ln^3(t)}{\sqrt{1-4t^2}}dt= \frac38\zeta(3) +\frac38\zeta(2)-\frac32 }}$$