# Integration Challenge

#### Shobhit

##### Member
Here is an interesting integral, which I would like to share with you:

Show that

\begin{align*} \int_0^{\frac{\pi}{2}}\sin^{-1}\left( \frac{\sin x}{\phi}\right) dx&= \frac{\pi^2}{12}-\frac{3}{4}\log^2 \phi \end{align*}

where $\phi$ is the Golden Ratio.

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#### Random Variable

##### Well-known member
MHB Math Helper
Let $\displaystyle I(a) = \int_{0}^{\pi/2} \arcsin (a \sin x) \ dx$ for $|a|<1$.

Then $\displaystyle I'(a) = \int_{0}^{\pi/2}\frac{\sin x}{\sqrt{1-a^{2} \sin^{2} x}} \ dx = \int_{0}^{\pi/2} \sin x \sum_{k=0}^{\infty} \binom{1/2+k-1}{k} (a \sin x)^{2k}$

$= \displaystyle \sum_{k=0}^{\infty} \frac{\Gamma(k+1/2)}{\Gamma(k+1) \Gamma(1/2)} a^{2k} \int_{0}^{\pi/2} \sin^{2k+1} x \ dx = \sum_{k=0}^{\infty} \frac{\Gamma(k+1/2)}{\Gamma(k+1) \Gamma(1/2)} a^{2k} \frac{2^{2k}}{2k+1} \frac{(k!)^{2}}{(2k)!}$

$\displaystyle = \sum_{k=0}^{\infty} \frac{\Gamma(2k) \Gamma(1/2) }{2^{2k-1} \Gamma(k) \Gamma(k+1) \Gamma(1/2)} a^{2k} \frac{2^{2k}}{2k+1} \frac{(k!)^{2}}{(2k)!} \frac{k}{k} = \sum_{k=0}^{\infty} \frac{(2k)!}{(k!)^{2}} a^{2k} \frac{1}{2k+1} \frac{(k!)^{2}}{(2k)!}$

$= \displaystyle \sum_{k=0}^{\infty} \frac{a^{2k}}{2k+1} = \frac{\text{arctanh} \ a}{a}$

And $\displaystyle I(a) = \int \frac{\text{arctanh} \ a}{a} \ da = \frac{1}{2} \int \frac{\ln(1+a)-\ln(1-a)}{a} \ da = \frac{1}{2} \Big( -\text{Li}_{2}(-a) + \text{Li}_{2}(a) \Big) + C$

where the constant of integration is zero

So $\displaystyle \int_{0}^{\pi /2} \arcsin \left( \frac{\sin x}{\varphi} \right) \ dx= \frac{1}{2} \Big( -\text{Li}_{2} \left(-\frac{1}{\varphi} \right) + \text{Li}_{2} \left( \frac{1}{\varphi} \right) \Big) = \frac{1}{2} \Big( \frac{\pi^{2}}{15} - \frac{\ln^{2} \varphi}{2} + \frac{\pi^{2}}{10} - \ln^{2} \varphi \Big)= \frac{\pi^{2}}{12} - \frac{3 \ln^{2} \varphi}{4}$

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#### Shobhit

##### Member
Very good RV! #### DreamWeaver

##### Well-known member
Nicely done, RV! #### Random Variable

##### Well-known member
MHB Math Helper
I didn't have to differentiate inside of the integral.

I could have used the Taylor expansion of $\arcsin x$ at $x=0$.

$\displaystyle I(a) = \int_{0}^{\pi/2} \arcsin (a \sin x) \ dx = \int_{0}^{\pi /2} \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^{2} 2^{2n} (2n+1)} \ (a \sin x)^{2n+1} \ dx$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^{2} 4^{n} (2n+1)} a^{2n+1} \int_{0}^{\pi /2} \sin^{2n+1} x \ dx = \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^{2} 2^{2n} (2n+1)} a^{2n+1} \frac{2^{2n}}{2n+1} \frac{(n!)^{2}}{(2n)!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{a^{2n+1}}{(2n+1)^{2}} = \chi_{2} (a) = \frac{1}{2} \Big( \text{Li}_{2}(a) - \text{Li}_{2} (-a) \Big)$

#### Random Variable

##### Well-known member
MHB Math Helper
You can do something similar for $\displaystyle \int_{0}^{\pi /2} \arcsin^{2} (a \sin x) \ dx$ by using the Taylor expansion of $\arcsin^{2}(x)$ at $x=0$

You'll get that $\displaystyle \int_{0}^{\pi /2} \arcsin^{2} (a \sin x) \ dx = \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(a^{2})^{n}}{n^{2}} = \frac{\pi}{4} \text{Li}_{2}(a^{2})$

Then, for example, $\displaystyle \int_{0}^{\pi /2} \arcsin^{2} \left( \frac{\sin x}{\sqrt{2}} \right) \ dx = \frac{\pi}{4} \text{Li}_{2} \left( \frac{1}{2} \right) = \frac{\pi^{3}}{48} - \frac{\pi}{8} \log^{2} 2$