primitive of a function with $e$ and trigonometric functions

Question

I have to find $$ \int_0^\frac{\pi}{2}\:e^{\sqrt{\cos x}}\sqrt{\sin2x}dx$$

Any help would be immensely appreciated.

What should I do?

Answer 1

It’s just an idea and we can try together to find a solution. Consider

$$\frac{d}{dx}\left(e^{\sqrt{\cos \left(x\right)}}\right)=-\frac{e^{\sqrt{\cos \left(x\right)}}\sin \left(x\right)}{2\sqrt{\cos \left(x\right)}}=-\frac{\sqrt{2}}{2}\frac{\sqrt{\sin \left(x\right)}}{\cos(x)}{e^{\sqrt{\cos \left(x\right)}}\sqrt{\sin \left(2x\right)}}$$

This means that your integral $I$ is equal to

$I=-2\sqrt{2} \int \frac{d}{dx}\left( \sqrt{\sin(x)}\right) \frac{d}{dx}\left(e^{\sqrt{\cos \left(x\right)}}\right) dx $

Now impose $y= e^{\sqrt{\cos \left(x\right)}}$. In this case

$$\frac{\log^2(y)}{2\sqrt[4]{1-\log^4(y)}}= \frac{d}{dx}\left( \sqrt{\sin(x)} \right)$$

and the integral become

$$I= -\sqrt{2} \int \frac{\log^2(y)}{\sqrt[4]{1-\log^4(y)}} dy$$

Impose $y=e^z$ to get

$$I= -\sqrt{2} \int \frac{z^2}{\sqrt[4]{1-z^4}} e^zdz$$

I think it’s very difficult to find a primitive of this function.

Answer 2

If you write $$e^{\sqrt{\cos (x)}}=\sum_{n=0}^\infty \frac {\cos^{\frac n 2}(x)}{n!} $$ you need to compute integrals $$I_n=\sqrt 2\int \sqrt{\sin (x)} \cos ^{\frac{n+1}{2}}(x)\,dx$$ $$x=\sin ^{-1}\left(t^2\right)\implies I_n=2\sqrt 2\int t^2 \,\left(1-t^4\right)^{\frac{n-1}{4}}\,dt$$ and now we need the gaussian hypergeometric function $$I_n=\frac{2\sqrt{2}}{3} \, t^3 \,\, _2F_1\left(\frac{3}{4},\frac{1-n}{4};\frac{7}{4};t^4\right)$$ Integrating between $t=0$ and $t=1$ gives $$I_n=\frac{2\sqrt{2}}{3}\,\Gamma \left(\frac{7}{4}\right) \,\frac{ \Gamma \left(\frac{n+3}{4}\right)}{\Gamma \left(\frac{n+6}{4}\right)}$$ This makes for the definite integral $$\int_0^\frac{\pi}{2}\:e^{\sqrt{\cos (x)}}\sqrt{\sin(2x)}\,dx=\frac{2\sqrt{2}}{3} \,\Gamma \left(\frac{7}{4}\right)\,\sum_{n=0}^\infty\frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma (n+1)\, \Gamma \left(\frac{n+6}{4}\right)}$$ The summation has a closed form (have a look here) in terms of generalized hypergeometric functions. At the end, using Wolfram Alpha results this gives for the definite integral $$I=\int_0^\frac \pi 2 e^{\sqrt{\cos (x)}}\,\sqrt{\sin (2 x)} \, dx=\frac{\pi}{8} \, _0F_3\left(;\frac{3}{4},\frac{3}{2},2;\frac{1}{4^4}\right)+\frac{2\sqrt{2}}{3} \, _1F_4\left(1;\frac{1}{2},\frac{3}{4},\frac{5}{4},\frac{7}{4};\frac{1}{4^4}\right )+$$ $$\frac{8\,\sqrt{\pi }\, \Gamma \left(\frac{7}{4}\right)}{45 \,\Gamma \left(\frac{1}{4}\right)} \left(15 \, _0F_3\left(;\frac{1}{4},\frac{1}{2},\frac{3}{2};\frac{1}{4^4}\right)+\sqrt{2} \, _0F_3\left(;\frac{5}{4},\frac{7}{4},\frac{9}{4};\frac{1}{4^4}\right)\right) $$

Numerically $$I=2.676843936995804519493015258301078853646728604573267\cdots$$

Even if this does not mean anything, this number is very close to the reciprocal of the first positive root of the quartic $$9589 x^4-3048 x^3-3616 x^2+235 x+389=0$$ The difference is $2.81\times 10^{-22}$.

Edit

We can have approximations of the definite integral since, using $x=\sin ^{-1}\left(t^2\right)$ $$I=\int_0^\frac \pi 2 e^{\sqrt{\cos (x)}}\,\sqrt{\sin (2 x)} \, dx=2 \sqrt 2\int_0^1 \frac{ t^2}{\sqrt[4]{1-t^4}}\,e^{\sqrt[4]{1-t^4}}\,dt$$

Expanded as a series $$\frac{ t^2}{\sqrt[4]{1-t^4}}\,e^{\sqrt[4]{1-t^4}}=e \Bigg[t^2+\frac{t^{10}}{32} \sum_{n=0}^\infty a_n\, t^{4n} \Bigg]$$ and the first coefficients $a_n$ are $$\left\{1,\frac{11}{12},\frac{13}{16},\frac{2789}{3840},\frac{12113}{18432},\frac{258539}{430080},\frac{6546397}{11796480},\cdots\right\}$$ The problem is that many terms would be required to have a decent approximation while

$$S_p=\sum_{n=0}^p\frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma (n+1)\, \Gamma \left(\frac{n+6}{4}\right)}$$ converges quite fast since $$\frac{a_{n+1}}{a_n}=\frac{\Gamma (n+1) \,\Gamma \left(\frac{n+4}{4}\right)\, \Gamma \left(\frac{n+6}{4}\right)}{\Gamma (n+2)\, \Gamma \left(\frac{n+3}{4}\right)\, \Gamma \left(\frac{n+7}{4}\right)}=\frac{1}{n}-\frac{7}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ Just to give an idea $$S_{12}= 3.089258529253\cdots$$ to be compared to the value $S_{\infty}=3.089258529315\cdots$ (difference of $6.23\times 10^{-11}$)