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Is there any closed formula for $\int_{0}^{\pi/2} \dfrac{e^{-x}\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}}?$

I know $\int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}} = \dfrac{\pi}{4},$ replacing $x$ by $\frac{\pi}{2} - y$.

amWhy
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Mathsource
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    $$\int_0^{\pi/2}\frac{e^{-x}\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}},dx=1-\int_0^\infty\frac{dx}{(1+x)^2e^{\arctan x^2}}$$ – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 13 '21 at 11:14
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    @TheSimpliFire, I don't think so that integral is enough to proceed. If you know how, then please let me know. – RAHUL Nov 13 '21 at 11:15
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    Wolfram can handle this with: $$\int_0^{\pi/2}e^{-x}\cdot\frac{\sqrt{\cos(x)}}{\sqrt{\cos(x)}+\sqrt{\sin(x)}},\mathrm{d}x=0.452394$$ But I am not certain if the decimal is an algebraic function of some polynomial in $\pi$. I set this up in GP-Pari and got $1001$ digits for the decimal, but attempting to find an inverse symbolic constant failed, due to the sites being down for indefinite maintenance. Integral value: 0.452393721949631630101048738406004050830342774266050312052872034252790383557926597398320787147141929148773232260394691326515298869430720510690674 – Randall Mar 05 '22 at 07:04
  • Using @Randall's result and a trigonometric inverse symbolic calculator, within an error of $1.5\times 10^{-9} $ $$\cos \left(\frac{27 \pi }{173}\right)-\sin \left(\frac{14 \pi }{99}\right)$$ – Claude Leibovici Aug 26 '22 at 07:43
  • I managed to achieve the following $$I=1-\frac{1}{1+e^{\pi/2}}\int_0^\infty\frac{\cosh(\operatorname{arctan}{(x^2)})}{(x+1)^2}$$ If that $x^2$ were an $x$, we would have an antiderivative with hypergeometric functions, except it's not. I don't know where to go from here – phi-rate Nov 30 '22 at 19:03
  • Similarly, to @TheSimpliFire we expand $e^{\tan^{-1}(x^2)}$ to get:$$I=1-(-1)^\frac i2\int_0^\infty \frac{\left(1-\frac{2i}{x^2+i}\right)^\frac i2}{(x+1)^2}dx$$ Also, this old question would not have satisfied the modern MSE quality standards, but it is a “historic” question – Тyma Gaidash Dec 06 '22 at 02:19
  • Also, here is a form of the integral only in terms of a product of $(x-k)^r$. Maybe there is a hypergeometric function for $\int_0^\infty \prod\limits_n (x-a_n)^{r_n}dx$? – Тyma Gaidash Dec 06 '22 at 02:35

2 Answers2

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$$I=\int_0^\frac{\pi}{2}\frac{e^{-\theta}\sqrt{\cos{\theta}}}{\sqrt{\sin{\theta}}+\sqrt{\cos{\theta}}}d\theta=\int_0^\frac{\pi}{2}\frac{e^{-\theta}}{1+\sqrt{\tan{\theta}}}d\theta$$ With integration by parts with $u=(1+\sqrt{\tan\theta})^{-1}$ and $dv=e^{-\theta}$ we get $$I=1-\int_0^\frac{\pi}{2}\frac{e^{-\theta} d(\sqrt{\tan{\theta}})}{(1+\sqrt{\tan\theta})^2}$$ Substitute $u=\sqrt{\tan\theta}$ $$I=1-\int_0^\infty\frac{e^{-\arctan(u^2)}}{(1+u)^2}dx=1+\int_0^\infty e^{-\arctan{(u^2)}}d\left(\frac{1}{1+u}\right)$$ Then substitute $v=(1+u)^{-1}$ $$I=1-\int_0^1\exp\left(-\arctan\left((v^{-1}-1)^2\right)\right)dv$$ Use the logarithmic definition of the arctangent $$I=1-\int_0^1\exp\left(\frac{i}{2}\ln\left(\frac{1+i(v^{-1}-1)^2}{1-i(v^{-1}-1)^2}\right)\right)dv$$$$=1-\int_0^1\left(\frac{1+i(v^{-1}-1)^2}{1-i(v^{-1}-1)^2}\right)^{i/2}dv$$$$=1-\int_0^1\left(\frac{(1+i)v^2-2iv+i}{(1-i)v^2+2iv-i}\right)^{i/2}dv$$$$=1-\left(\frac{1+i}{1-i}\right)^{i/2}\int_0^1\left(\frac{v^2+(-1-i)v+(1+i)/2}{v^2+(-1+i)v+(1-i)/2}\right)^{i/2}dv$$$$=1-i^{i/2}\int_0^1\left(1-\frac{(2v-1)i}{v^2+(-1+i)v+(1-i)/2}\right)^{i/2}$$ Now we can use the following series I derived to continue. We can easily check it with negative integer inputs for k, and we can use the gamma function to expand it to the complex numbers $$(1-x)^k=\sum_{n=0}^\infty\frac{\Gamma(n-k)}{\Gamma(-k)}\frac{x^n}{n!}, k≠1,2,3,4,...$$ $$I=1-e^{-\pi/4}\sum_{n=0}^\infty\frac{\Gamma(n-i/2)}{\Gamma(-i/2)}\frac{i^n}{n!}\int_0^1\left(\frac{2v-1}{v^2+(-1+i)v+(1-i)/2}\right)^ndv$$ We can use special functions to solve this integral. I asked a question for this here. I'll add to this later

Let's handle this integral separately. $$I_n=\int_0^1\left(\frac{2v-1}{v^2+(-1+i)v+(1-i)/2}\right)^ndv$$ We want to substitute $w=\frac{2v-1}{v^2+(-1+i)v+(1-i)/2}$, but we're going to need algebra first $$w=\frac{2v-1}{v^2+(-1+i)v+(1-i)/2}$$ $$wv^2+(-1+i)wv+\frac{1-i}{2}w=2v-1$$ $$wv^2+((-1+i)w-2)v+(\frac{1-i}{2}w+1)=0$$ $$v=\frac{-((-1+i)w-2)+\sqrt{((-1+i)w-2)^2-4(w)((\frac{1-i}{2}w+1))}}{2(w)}$$ $$v=\frac{2+(1-i)w+\sqrt{-2w^2-4iw+4}}{2w}$$ $$v=\frac{1}{w}+\frac{1-i}{2}+\frac{1}{2}\sqrt{-2-4iw^{-1}+4w^{-2}}$$ $$dv=\left(-\frac{1}{w^2}+\frac{iw^{-2}-2w^{-3}}{\sqrt{-2-4iw^{-1}+4w^{-2}}}\right)dw$$ $$dv=\left(-\frac{1}{w^2}+\frac{iw^{-2}-2w^{-3}}{\sqrt{-2-4iw^{-1}+4w^{-2}}}\right)dw$$ $$dv=\left(-\frac{1}{w^2}+\frac{iw^{-1}-2w^{-2}}{\sqrt{-2w^2-4iw+4}}\right)dw$$ Now that our substitution is prepared$$I_n=\int_{-1-i}^{-1+i}w^n\left(-\frac{1}{w^2}+\frac{iw^{-1}-2w^{-2}}{\sqrt{-2w^2-4iw+4}}\right)dw$$ Let $$J_n=\int_{-1-i}^{-1+i}\frac{w^n}{\sqrt{-2w^2-4iw+4}}dw$$ Then $$I_n=\frac{(-1-i)^{n-1}-(1-i)^{n-1}}{n+1}+iJ_{n-1}-2J_{n-2}$$ and $$I=1-e^{-\pi/4}\sum_{n=0}^\infty\frac{\Gamma(n-i/2)}{\Gamma(-i/2)}\frac{i^n}{n!}\left(\frac{(-1-i)^{n-1}-(1-i)^{n-1}}{n+1}+iJ_{n-1}-2J_{n-2}\right)$$ I'll add more to this later. I already have a solution to $J_n$

phi-rate
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Here is an experimental result using a special case of Lauricella D from the DLMF’s hyperelliptic $\text R$ function which has a multiple series expansion, using the Pochhammer symbol $(u)_v$, from

Is this Lauricella $\text F_\text D$ to hypergeometric R, from DLMF, conversion formula correct?:

$$\text B(-a,(-a)’)\text R_a(b_1,\dots,b_n;z_1,\dots z_n)=\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt=\sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(-a)_{\sum_{j=1}^n m_j}}{\left(\sum\limits_{j=1}^nb_j\right)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n\frac{(b_j)_{m_j}(1-x_j)^{m_j}}{m_j!},(-a)’=a+\sum_{j=1}^n b_j$$

Using

$$I=1-(-1)^\frac i2\int_0^\infty t^0(t-\sqrt i)^\frac i2(t+\sqrt i)^\frac i2(t-\sqrt{-i})^{-\frac i2}(t+\sqrt{-i})^{-\frac i2}(t+1)^{-2}dt$$

and $(-a)’=1=a-\frac i2-\frac i2+\frac i2+\frac i2+2\implies a=-1$

Therefore?:

$$I=1-i^i\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;-\sqrt i,\sqrt i,-\sqrt{-i},\sqrt {-i},1\right)$$

One problem is $x_5=1$ gives $(1-x_5)^{m_5}=0^{m_5}$ in the series expansion. Fortunately, the function is homogenous:

$$1-(i^{-i})^{-1}\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;-\sqrt i,\sqrt i,-\sqrt{-i},\sqrt {-i},1\right) = 1-\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;i^{\frac 52-i},i^{\frac12-i},i^{\frac32-i},i^{-\frac12-i},i^{-i}\right) $$

Now for the possible series expansion

Тyma Gaidash
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