After reading the original paper of Chowla-Selberg, I think it is a problem which can be solved with Kummer's Fourier Expansion of Gamma Function

$$\log\Gamma(x)=(\frac{1}{2}-x)(\gamma+\log2)+(1-x)\log\pi-(\log\sin\pi x)/2+\sum_{m=1}^{\infty}\frac{\sin 2\pi m x}{\pi m}\log m,$$

where $0<x<1$.

We do not need to worry about the first and the second term in Kummer's expansion because they cancel each other in addition. And it is not hard to show that $$\log\frac{\sin(\pi/24)\sin(11\pi/24)}{\sin(5\pi/24)\sin(7\pi/24)}=\log(2-\sqrt{3}).$$

We also need to figure out the sum $$S(m)={\sin(2\pi m/24)+\sin(2\pi m\times 11 /24)-\sin(2\pi m\times 5 /24)-\sin(2\pi m\times 7 /24)}.$$ It is nothing more than half of the Gauss sum for some Dirichlet characters $\chi$ modulo 24, which can be constructed like this:

$$\chi(1)=\chi(11)=\chi(17)=\chi(19)=1$$ and $$\chi(5)=\chi(7)=\chi(13)=\chi(23)=-1.$$

It is not hard to get $S(m)=0$ when $m$ is even, $S(m)=-\sqrt{2}\chi(m)$ when $m$ is prime to 24, and $S(3m)=2\sqrt{2}\chi_1(m)$ when $m$ is odd, where $\chi_1$ is a Dirichlet character modulo 8 and $$\chi_1(1)=\chi_1(3)=1,\chi_1(5)=\chi_1(7)=-1.$$ So we need to figure out the sum $$-\frac{\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\log n+\frac{2\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{3n}\log (3n)$$. We notice that $\chi(n)=\chi_1(n)$ when $n$ is prime to 24. So the sum is equal to $$\frac{\sqrt{2}\log 3}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}$$. We know that $$\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}=\frac{\pi}{2\sqrt{2}}$$ from class number formula, and we are done.

After reading the original paper of Chowla-Selberg, I think it is a problem which can be solved with Kummer's Fourier Expansion of Gamma Function

$$\log\Gamma(x)=(\frac{1}{2}-x)(\gamma+\log2)+(1-x)\log\pi-(\log\sin\pi x)/2+\sum_{m=1}^{\infty}\frac{\sin 2\pi m x}{\pi m}\log m,$$

where $0<x<1$.

We do not need to worry about the first and the second term in Kummer's expansion because they cancel each other in addition. And it is not hard to show that $$\log\frac{\sin(\pi/24)\sin(11\pi/24)}{\sin(5\pi/24)\sin(7\pi/24)}=\log(2-\sqrt{3}).$$

We also need to figure out the sum $$S(m)={\sin(2\pi m/24)+\sin(2\pi m\times 11 /24)-\sin(2\pi m\times 5 /24)-\sin(2\pi m\times 7 /24)}.$$ It is nothing more than half of the Gauss sum for some Dirichlet characters $\chi$ modulo 24, which can be constructed like this:

$$\chi(1)=\chi(11)=\chi(17)=\chi(19)=1$$ and $$\chi(5)=\chi(7)=\chi(13)=\chi(23)=-1.$$

It is not hard to get $S(m)=0$ when $m$ is even, $S(m)=-\sqrt{2}\chi(m)$ when $m$ is prime to 24, and $S(3m)=2\sqrt{2}\chi_1(m)$ when $m$ is odd, where $\chi_1$ is a Dirichlet character modulo 8 and $$\chi_1(1)=\chi_1(3)=1,\chi_1(5)=\chi_1(7)=-1.$$ So we need to figure out the sum $$-\frac{\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\log n+\frac{2\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{3n}\log (3n).$$ We notice that $\chi(n)=\chi_1(n)$ when $n$ is prime to 24. So the sum is equal to $$\frac{\sqrt{2}\log 3}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}.$$ We know that $$\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}=\frac{\pi}{2\sqrt{2}}$$ from class number formula, and we are done.

After reading the original paper of Chowla-Selberg, I think it is a problem which can be solved with Kummer's Fourier Expansion of Gamma Function

$$\log\Gamma(x)=(\frac{1}{2}-x)(\gamma+\log2)+(1-x)\log\pi-(\log\sin\pi x)/2+\sum_{m=1}^{\infty}\frac{\sin 2\pi m x}{\pi m}\log m,$$

where $0<x<1$.

We do not need to worry about the first and the second term in Kummer's expansion because they cancel each other in addition. And it is not hard to show that $$\log\frac{\sin(\pi/24)\sin(11\pi/24)}{\sin(5\pi/24)\sin(7\pi/24)}=\log(2-\sqrt{3}).$$

We also need to figure out the sum $$S(m)={\sin(2\pi m/24)+\sin(2\pi m\times 11 m/24)-\sin(2\pi m\times 5 /24)-\sin(2\pi m\times 7 /24)}.$$ It is nothing more than half of the Gauss sum for some Dirichlet characters $\chi$ modulo 24, which can be constructed like this:

$$\chi(1)=\chi(11)=\chi(17)=\chi(19)=1$$ and $$\chi(5)=\chi(7)=\chi(13)=\chi(23)=-1.$$

It is not hard to get $S(m)=0$ when $m$ is even, $S(m)=-\sqrt{2}\chi(m)$ when $m$ is prime to 24, and $S(m)=2\sqrt{2}\chi_1(m)$ where $\chi_1$ is a Dirichlet character modulo 8 and $$\chi_1(1)=\chi_1(3)=1,\chi_1(5)=\chi_1(7)=-1$$. So we need to figure out the sum $$-\frac{\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\log n+\frac{2\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{3n}\log (3n)$$. We notice that $\chi(n)=\chi_1(n)$ when $n$ is prime to 24. So the sum is equal to $$\frac{\sqrt{2}\log 3}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}$$. We know that $$\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}=\frac{\pi}{2\sqrt{2}}$$ from class number formula, and we are done.

After reading the original paper of Chowla-Selberg, I think it is a problem which can be solved with Kummer's Fourier Expansion of Gamma Function

$$\log\Gamma(x)=(\frac{1}{2}-x)(\gamma+\log2)+(1-x)\log\pi-(\log\sin\pi x)/2+\sum_{m=1}^{\infty}\frac{\sin 2\pi m x}{\pi m}\log m,$$

where $0<x<1$.

We do not need to worry about the first and the second term in Kummer's expansion because they cancel each other in addition. And it is not hard to show that $$\log\frac{\sin(\pi/24)\sin(11\pi/24)}{\sin(5\pi/24)\sin(7\pi/24)}=\log(2-\sqrt{3}).$$

We also need to figure out the sum $$S(m)={\sin(2\pi m/24)+\sin(2\pi m\times 11 /24)-\sin(2\pi m\times 5 /24)-\sin(2\pi m\times 7 /24)}.$$ It is nothing more than half of the Gauss sum for some Dirichlet characters $\chi$ modulo 24, which can be constructed like this:

$$\chi(1)=\chi(11)=\chi(17)=\chi(19)=1$$ and $$\chi(5)=\chi(7)=\chi(13)=\chi(23)=-1.$$

It is not hard to get $S(m)=0$ when $m$ is even, $S(m)=-\sqrt{2}\chi(m)$ when $m$ is prime to 24, and $S(3m)=2\sqrt{2}\chi_1(m)$ when $m$ is odd, where $\chi_1$ is a Dirichlet character modulo 8 and $$\chi_1(1)=\chi_1(3)=1,\chi_1(5)=\chi_1(7)=-1.$$ So we need to figure out the sum $$-\frac{\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\log n+\frac{2\sqrt{2}}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{3n}\log (3n)$$. We notice that $\chi(n)=\chi_1(n)$ when $n$ is prime to 24. So the sum is equal to $$\frac{\sqrt{2}\log 3}{\pi}\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}$$. We know that $$\sum_{n=1}^{\infty}\frac{\chi_1(n)}{n}=\frac{\pi}{2\sqrt{2}}$$ from class number formula, and we are done.