compute the following integral in closed form : $\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2}(x)+8\cos^{2} x}dx$

Question

Evaluate

$I=\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2} (x)+8\cos^{2} x}dx$ How can I starte in this hard integral , at first use $y=\frac{π}{2}$ but no result so I this use :

$y=\tan \frac{x}{2}$ then $dx=2\frac{1}{1+y^2}dy$

$x=2\arctan y$

$\cos x=\frac{1-y^2}{1+y^2}$ $&$ $\sin x=2\frac{y}{1+y^2}$

So :

$8\cos^{2} x+(1+\sqrt 2)sin^{2} x=\frac{8(1-y^2)^2+4(1+\sqrt 2)y^2}{(1+y^2)^2}$

Now I get $arctan$ integral

$I=2\int_0^{\infty}\frac{(1+y^2)\arctan y}{8(1-y^2)^2+4(1+\sqrt 2)y^2}dy$

But I don't know how to complete this work!

Answer 1

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Define the function $\mathcal{I}:\left(0,\infty\right)\rightarrow\mathbb{R}$ via the definite integral

$$\begin{align} \mathcal{I}{\left(z\right)} &:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\vartheta}{z^{2}\sin^{2}{\left(\vartheta\right)}+\cos^{2}{\left(\vartheta\right)}}.\\ \end{align}$$

Then, the value of $I$ can be computed as

$$I=\frac18\,\mathcal{I}{\left(\sqrt{\frac{1+\sqrt{2}}{8}}\right)}.$$

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Suppose $z\in\left(0,1\right)$. Then $0<\frac{1}{z}-z$, and we find

$$\begin{align} \mathcal{I}{\left(z\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\vartheta}{z^{2}\sin^{2}{\left(\vartheta\right)}+\cos^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\vartheta}{1-\left(1-z^{2}\right)\sin^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\vartheta\int_{0}^{\vartheta}\mathrm{d}\varphi\,\frac{1}{1-\left(1-z^{2}\right)\sin^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{1}{1-\left(1-z^{2}\right)\sin^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{1}{1-\left(1-z^{2}\right)\left[1-\cos^{2}{\left(\vartheta\right)}\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\sec^{2}{\left(\vartheta\right)}}{\sec^{2}{\left(\vartheta\right)}-\left(1-z^{2}\right)\left[\sec^{2}{\left(\vartheta\right)}-1\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\sec^{2}{\left(\vartheta\right)}}{1+\tan^{2}{\left(\vartheta\right)}-\left(1-z^{2}\right)\tan^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\varphi}^{\frac{\pi}{2}}\mathrm{d}\vartheta\,\frac{\sec^{2}{\left(\vartheta\right)}}{1+z^{2}\tan^{2}{\left(\vartheta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{\tan{\left(\varphi\right)}}^{\infty}\mathrm{d}t\,\frac{1}{1+z^{2}t^{2}};~~~\small{\left[\vartheta=\arctan{\left(t\right)}\right]}\\ &=\frac{1}{z}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{z\tan{\left(\varphi\right)}}^{\infty}\mathrm{d}x\,\frac{1}{1+x^{2}};~~~\small{\left[t=\frac{x}{z}\right]}\\ &=\frac{1}{z}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\left[\frac{\pi}{2}-\arctan{\left(z\tan{\left(\varphi\right)}\right)}\right]\\ &=\frac{1}{z}\int_{0}^{\infty}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(zt\right)}\right]}{1+t^{2}};~~~\small{\left[\varphi=\arctan{\left(t\right)}\right]}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(zt\right)}\right]}{1+t^{2}}+\frac{1}{z}\int_{1}^{\infty}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(zt\right)}\right]}{1+t^{2}}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(zt\right)}\right]}{1+t^{2}}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(\frac{z}{t}\right)}\right]}{1+t^{2}};~~~\small{\left[t\mapsto\frac{1}{t}\right]}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}-\arctan{\left(zt\right)}\right]}{1+t^{2}}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{t}{z}\right)}}{1+t^{2}}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}+\arctan{\left(\frac{t}{z}\right)}-\arctan{\left(zt\right)}\right]}{1+t^{2}}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}+\arctan{\left(\frac{\left(\frac{t}{z}\right)-\left(zt\right)}{1+\left(\frac{t}{z}\right)\left(zt\right)}\right)}\right]}{1+t^{2}}\\ &=\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\left[\frac{\pi}{2}+\arctan{\left(\frac{\left(\frac{1}{z}-z\right)t}{1+t^{2}}\right)}\right]}{1+t^{2}}\\ &=\frac{\pi}{2z}\int_{0}^{1}\mathrm{d}t\,\frac{1}{1+t^{2}}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{\left(\frac{1}{z}-z\right)t}{1+t^{2}}\right)}}{1+t^{2}}\\ &=\frac{\pi^{2}}{8z}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{\left(\frac{1}{z}-z\right)t}{1+t^{2}}\right)}}{1+t^{2}}.\\ \end{align}$$

Setting $r:=\frac{1}{2z}-\frac{z}{2}\in\mathbb{R}_{>0}$, we then obtain the following closed-form expression in terms of the Legendre chi function using one of the integral representations at the bottom of the page:

$$\begin{align} \mathcal{I}{\left(z\right)} &=\frac{\pi^{2}}{8z}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{\left(\frac{1}{z}-z\right)t}{1+t^{2}}\right)}}{1+t^{2}}\\ &=\frac{\pi^{2}}{8z}+\frac{1}{z}\int_{0}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{2rt}{1+t^{2}}\right)}}{1+t^{2}}\\ &=\frac{\pi^{2}}{8z}+\frac{1}{2z}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\arctan{\left(r\sin{\left(\varphi\right)}\right)};~~~\small{\left[t=\tan{\left(\frac{\varphi}{2}\right)}\right]}\\ &=\frac{\pi^{2}}{8z}+\frac{1}{z}\,\chi_{2}{\left(\frac{\sqrt{1+r^{2}}-1}{r}\right)}\\ &=\frac{\pi^{2}}{8z}+\frac{1}{z}\,\chi_{2}{\left(\frac{1-z}{1+z}\right)}.\blacksquare\\ \end{align}$$

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Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi/2}\frac{x}{\pars{1 + \root{2}} \sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x:\ {\LARGE ?}}$.

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi/2}{x \over \pars{1 + \root{2}} \sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x} \\[5mm] & = \int_{0}^{\pi/2}{x \over \pars{1 + \root{2}} \bracks{1 - \cos\pars{2x}}/2 + 8\bracks{1 + \cos\pars{2x}}/2}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\pi/2}{2x \over 9 + \root{2} + \pars{7 - \root{2}} \cos\pars{2x}}\,2\,\dd x \\[5mm] & = {7 + \root{2} \over 94} \int_{0}^{\pi}{x \over a + \cos\pars{x}}\,\dd x \end{align} where $\ds{a \equiv {65 + 16\root{2} \over 47} > 1}$.

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Then, \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\pi}{x \over a + \cos\pars{x}}\,\dd x} \\[5mm] = &\ \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi}{-\ic\ln\pars{z} \over a + \pars{z + 1/z}/2}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left. -2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi}{\ln\pars{z} \over z^{2} + 2az + 1}\,\dd z \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \tag{1}\label{1} \end{align} Roots of $\ds{z^{2} + 2az + 1 = 0}$ are given by $\ds{r_{\pm} \equiv -a \pm \root{a^{2} - 1}}$ where $\ds{r_{-} < -1}$ and $\ds{-1 < r_{+} < 0}$.

\eqref{1} can be handle with the "Polylogarithm Machinery".

Answer 3

Let $a=\sqrt{8(\sqrt2-1)}$\begin{align} &\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2} (x)+8\cos^{2} x}dx\\ \overset{ibp}=& \ \frac a8\int_0^{\frac{π}{2}}\cot^{-1}\frac{\tan x}{a}dx = \frac a8 \int^\infty_{\frac{1}{a}}\frac{\ln t}{t^2-1}dt \\ =& \ \frac a{16}\bigg(\frac{\pi^2}3 +\text{Li}_2(\frac{a-1}a)+ \text{Li}_2(\frac{1}a)-\ln a \ln\frac{a+1}a\bigg)\\ \end{align}

Answer 4

Consider $$I := \int_0^{\frac\pi2} \frac{x \,dx}{\alpha^2 \sin^2 x + \beta^2 \cos^2 x}.$$ Changing variables to $$u = \tan x$$ transforms the integral to $$\int_0^\infty \frac{\arctan u \,du}{\alpha^2 u^2 + \beta^2}.$$ Now, consider the family $$I(\lambda) := \int_0^\infty \frac{\arctan \lambda u \,du}{\alpha^2 u^2 + \beta^2}.$$ of integrals; in particular, $I(1) = I$. Differentiating under the integral sign gives $$I'(\lambda) = \int_0^\infty \frac{du}{(\lambda^2 u^2 + 1) (\alpha^2 u^2 + \beta^2)} = \frac{\log \frac\beta\alpha + \log \lambda}{\beta \lambda^2 - \alpha^2} .$$ Now, $I(\infty) := \lim_{\lambda \to \infty} I(\lambda) = \frac\pi2 \int_0^\infty \frac{du}{\alpha^2 u^2 + \beta} = \frac{\pi^2}{4 \alpha \beta}$, so $$ I(\infty) - I = \frac{\pi^2}{4 \alpha \beta} - I .$$ On the other hand, the F.T.C. gives $$I(\infty) - I = \int_1^\infty \frac{\log \frac\beta\alpha + \log \lambda}{\beta^2 \lambda^2 - \alpha^2} \,d\lambda$$ and evaluating this integral, e.g., using the identity $$\int_0^1 \frac{\log y \,dy}{a^2 - y^2} = \frac{1}{2a} \left(\operatorname{Li}_2\left(-\frac1a\right) - \operatorname{Li}_2\left(\frac1a\right)\right) ,$$ and rearranging yields $$I = \frac1{2 \alpha \beta} \left(-2 \log \zeta \operatorname{artanh} \zeta + \operatorname{Li}_2\left(\zeta\right) - \operatorname{Li}_2\left(-\zeta\right) \right), $$ where $\zeta := \frac\beta\alpha$. The particular values $\alpha = \sqrt{1 + \sqrt 2}, \beta = \sqrt 8$ do not appear to simplify the expression. Evaluating numerically for those $\alpha, \beta$ gives $I = 0.3475504241\ldots$.