Generalized log integral $\int_{0}^{\frac{\pi}{2}}\log(a+\sin(x))\,dx$

Question

I was trying to evaluate the following integral, and wanted a solution verification. If I did something wrong please explain why it was wrong and how it can be fixed!

Here is my method:

$$I:=\int_{0}^{\frac{\pi}{2}}\log(a+\sin(x))\,dx$$

$$=\int_{0}^{\frac{\pi}{2}}\log(a+\cos(x))\,dx$$

Use:

$$\cos(x)\equiv 2\cos^2(\frac{x}{2})-1$$

And define:

$$c:=a-1$$

So

$$I=\int_{0}^{\frac{\pi}{2}}\log(c+2\cos^2(\frac{x}{2}))\,dx$$

Let

$$\frac{x}{2}\longrightarrow{x}$$

$$I=2\int_{0}^{\frac{\pi}{4}}\log(c+2\cos^2(x))\,dx$$

$$=2\int_{0}^{\frac{\pi}{4}}\log(c\sec^2(x)+2)\,dx+4\int_{0}^{\frac{\pi}{4}}\log(\cos(x))\,dx$$

The second integral is evaluated by my post:

Solving integrals through “integral systems.”

For the first, let:

$$\tan(x)\longrightarrow{x}$$

$$I=2G-\pi\log(2)+2\int_{0}^{1}\frac{\log(c(x^2+1)+2)}{x^2+1}\,dx$$

$$I=2G-\pi\log(2)+2J$$

$$J=J(c):=\int_{0}^{1}\frac{\log(c(x^2+1)+2)}{x^2+1}\,dx$$

Now we perform Feynman’s technique on $J(c)$.

$$J’(c)=\int_{0}^{1}\frac{1}{cx^2+c+2}\,dx$$

$$J’(c)=\frac{\tan^{-1}\sqrt{\frac{c}{2+c}}}{\sqrt{c(2+c)}}$$

$$J(c)=J(c)-J(0)+\frac{\pi}{4}\log(2)=\frac{\pi}{4}\log(2)+\int_{0}^{c}J’(x)\,dx$$

$$J=\frac{\pi}{4}\log(2)+\int_{0}^{c}\frac{\tan^{-1}\sqrt{\frac{x}{2+x}}}{\sqrt{x(2+x)}}\,dx$$

Let:

$$\sqrt{\frac{x}{2+x}}\longrightarrow{x}$$

$$J=\frac{\pi}{4}\log(2)+2\int_{0}^{\sqrt{\frac{a-1}{a+1}}}\frac{\tan^{-1}(x)}{1-x^2}\,dx$$

By another one of my posts:

A generalized integral. $\int_{0}^{t}\frac{\arctan(x)}{1-x^2}\,dx$

And with mulch simplification we arrive at:

$$I=\frac{\pi}{2}\cosh^{-1}(a)-\frac{\pi}{2}\log(2)+2Ti_2(a-\sqrt{a^2-1})$$

Where in my computations, $G$ denotes Catalans constant, and $Ti_2(x)$ denotes the inverse tangent integral.

Answer 1

Your work seems correct to me. I actually like your approach a lot. You may however want to define exactly what $a$ is. Here is another solution using my favorite methods.

$$\begin{align*}I(a)&=\int_{0}^{\pi/2}\log(a+\sin(x))\ dx \\&=\int_{0}^{\pi/2}\log(a)+\log\left(1+\frac{\sin(x)}{a}\right)\ dx \\&=\frac{\pi\log(a)}{2}+\underbrace{\int_{0}^{\pi/2}\log\left(1+\frac{\sin(x)}{a}\right)\ dx}_{J(a)}\end{align*}$$ Expanding the logarithm by it's power series, $$J(a)=\int_0^{\pi/2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\frac{\sin^n(x)}{a^n}\ dx=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}\int_0^{\pi/2}\sin^{n}(x)\ dx$$ by the integral representation of the Beta function, $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}\int_0^{\pi/2}\sin^{n}(x)\ dx=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}B\left(\frac{n+1}{2},\frac{1}{2}\right)$$ which in terms of the Gamma function, $$\begin{align*}\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}B\left(n+\frac{1}{2},\frac{1}{2}\right)&=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)} \\ &=\frac{\sqrt\pi}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{na^n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}\end{align*}$$ then by the relation, $$\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}=\frac{\sqrt\pi}{2^n}\binom{n}{n/2}$$ we have our series, $$J(a)=-\frac{\pi}{2}\sum_{n=1}^\infty\binom{n}{n/2}\frac{(-1)^{n}}{n(2a)^n}$$ which is evaluated here (credits to @ClaudeLeibovici) and offers closed forms in terms of dilogarithms and hyperbolic functions, but it is not pretty.

Instead if you allow the ${}_3F_2$ Hypergeometric function the integral is, $$I(a)=-\frac{\pi}{2}\log\left(2\left(a-\sqrt{a^2-1}\right)\right)+\frac{1}{a}{}_3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};\frac{1}{a^2}\right).$$

Answer 2

Considering the parameterized integral $$ I(\theta)=\int_0^\pi \ln (1+\cos \theta \cos x) d x $$ Differentiating $I(\theta)$ w.r.t. $\theta$ yields $$ \begin{aligned} I^{\prime}(\theta)&= -\int_0^{\frac{\pi}{2}} \frac{\sin \theta \cos x}{1+\cos \theta \cos x} d x\\ &= -\frac{\sin \theta}{\cos \theta} \int_0^{\frac{\pi}{2}} \frac{1+\sec \theta \cos x-1}{1+\cos \theta \cos x} d x \\ & =-\frac{\pi \tan \theta}{2}+\tan \theta \int_0^{\frac{\pi}{2}} \frac{d x}{1+\cos \theta \cos x} \\ & =-\frac{\pi\tan \theta}{2}+\tan \theta \cdot \frac{\theta}{\sin \theta} \cdots (*) \\ & =-\frac{\pi \tan \theta}{2}+\frac{\theta}{\cos \theta} \\ & \end{aligned} $$ where $(*)$ refer to the post.

Integrating back from $0$ to $\theta$ gives $$ \begin{aligned} I(\theta)-I(0)&=\int_0^\theta\left(-\frac{\pi \tan t}{2}+\frac{t}{\cos t}\right) d t \\ & =\frac{\pi}{2} \ln (\cos \theta)+\int_0^\theta \frac{t}{\cos t} d t \end{aligned} $$ Using $I(0)=\int_0^{\frac{\pi}{2}} \ln (1+\cos x) d x=2G-\frac{\pi}{2} \ln 2$, where $G$ is the Catalan’s constant, we have $$ \int_0^{\frac{\pi}{2}} \ln (1+\cos \theta \cos x) d x= 2 G-\frac{\pi}{2} \ln 2+\frac{\pi}{2} \ln (\cos \theta)+\int_0^\theta \frac{t}{\cos t} d t $$

By the post, we have

$$\int\frac{t}{\cos t}=-2it\arctan(e^{it})+i(\mathrm{Li}_2(-ie^{it})-\mathrm{Li}_2(ie^{it}))$$ and hence $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \ln (1+\cos \theta \cos x) d x& =2G-\frac{\pi}{2} \ln 2+\frac{\pi}{2} \ln (\cos \theta)-2 i \theta \tan^{-1}\left(e^{\theta i}\right) \\ & \quad +i\left(\mathrm{Li}_2\left(-i e^{\theta i}\right)\right)-\mathrm{Li}_2\left(i e^{i \theta}\right) +i\mathrm{Li}_2(-i)-\mathrm{Li}_2(i) \end{aligned} $$ Come back to our integral for $a>1$, we have

$$\int_0^{\frac{\pi}{2}} \ln (a+\sin x) d x=\int_0^{\frac{\pi}{2}} \ln (a+\cos x) d x = \frac{\pi}{2} \ln a+\int_0^{\frac{\pi}{2}} \ln \left(1+\frac{1}{a} \cos x\right) d x $$ Letting $a=\sec \theta$ brings us

$$ \int_0^{\frac{\pi}{2}} \ln (a+\sin x) d x = 2G+\frac{\pi}{2} \ln \frac {a}{2}-2 i\sec^{-1} a\tan^{-1}\left(e^{i\sec^{-1} a }\right)+i\left(\mathrm{Li}_2\left(-i e^{i\sec^{-1} a}\right)\right)-\mathrm{Li}_2\left(i e^{i\sec^{-1} a}\right) +i\mathrm{Li}_2(-i)-\mathrm{Li}_2(i) $$

Answer 3

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Define the function $\mathcal{I}:\mathbb{R}_{\ge0}\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(a\right)}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}.$$

The $a\ge1$ case has already been addressed elsewhere, so this answer will just consider the $0<a<1$ case.

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Given $a\in(0,1)$ and setting $\arcsin{\left(a\right)}=:\alpha\in\left(0,\frac{\pi}{2}\right)$, we find

$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(\sin{\left(\alpha\right)}+\sin{\left(\varphi\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\alpha+\varphi}{2}\right)}\cos{\left(\frac{\alpha-\varphi}{2}\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\left[-\ln{\left(2\right)}+\ln{\left(2\sin{\left(\frac{\alpha+\varphi}{2}\right)}\right)}+\ln{\left(2\cos{\left(\frac{\alpha-\varphi}{2}\right)}\right)}\right]\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi+\alpha}{2}\right)}\right)}+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\frac{\varphi-\alpha}{2}\right)}\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\frac{\pi}{2}+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)};~~~\small{\left[\varphi=\vartheta-\alpha\right]}\\ &~~~~~+\int_{-\alpha}^{\frac{\pi}{2}-\alpha}\mathrm{d}\omega\,\ln{\left(2\cos{\left(\frac{\omega}{2}\right)}\right)};~~~\small{\left[\varphi=\omega+\alpha\right]}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\frac{\pi}{2}+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)}\\ &~~~~~+\int_{\frac{\pi}{2}+\alpha}^{\pi+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)};~~~\small{\left[\omega=\pi-\vartheta\right]}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\int_{\alpha}^{\pi+\alpha}\mathrm{d}\vartheta\,\ln{\left(2\sin{\left(\frac{\vartheta}{2}\right)}\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi+\alpha\right)},\\ \end{align}$$

where the Clausen function is given by the integral

$$\operatorname{Cl}_{2}{\left(\theta\right)}=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|};~~~\small{\theta\in\mathbb{R}}.$$

Basic properties of the Clausen function include:

$$\operatorname{Cl}_{2}{\left(m\pi\right)}=0;~~~\small{m\in\mathbb{Z}},$$

$$\operatorname{Cl}_{2}{\left(-\theta\right)}=-\operatorname{Cl}_{2}{\left(\theta\right)};~~~\small{\theta\in\mathbb{R}},$$

$$\operatorname{Cl}_{2}{\left(\theta+2m\pi\right)}=\operatorname{Cl}_{2}{\left(\theta\right)};~~~\small{m\in\mathbb{Z}\land\theta\in\mathbb{R}}.$$

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Recall the definition of the so-called inverse tangent integral function:

$$\operatorname{Ti}_{2}{\left(x\right)}:=\int_{0}^{x}\mathrm{d}t\,\frac{\arctan{\left(t\right)}}{t};~~~\small{x\in\mathbb{R}}.$$

For $\theta\in\left(0,\frac{\pi}{2}\right)$, the inverse tangent integral $\operatorname{Ti}_{2}{\left(\tan{\left(\theta\right)}\right)}$ has the following closed form in terms of Clausen functions:

$$\begin{align} \operatorname{Ti}_{2}{\left(\tan{\left(\theta\right)}\right)} &=\int_{0}^{\tan{\left(\theta\right)}}\mathrm{d}t\,\frac{\arctan{\left(t\right)}}{t}\\ &=\int_{0}^{\theta}\mathrm{d}\vartheta\,\frac{\vartheta\sec^{2}{\left(\vartheta\right)}}{\tan{\left(\vartheta\right)}};~~~\small{\left[\arctan{\left(t\right)}=\vartheta\right]}\\ &=\int_{0}^{\theta}\mathrm{d}\vartheta\,\vartheta\csc{\left(\vartheta\right)}\sec{\left(\vartheta\right)}\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\int_{0}^{\theta}\mathrm{d}\vartheta\,\ln{\left(\tan{\left(\vartheta\right)}\right)};~~~\small{I.B.P.}\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\int_{0}^{2\theta}\mathrm{d}\varphi\,\frac12\ln{\left(\tan{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\vartheta=\frac{\varphi}{2}\right]}\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}-\frac12\int_{0}^{2\theta}\mathrm{d}\varphi\,\left[\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}-\ln{\left(2\cos{\left(\frac{\varphi}{2}\right)}\right)}\right]\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[-\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\frac{\varphi}{2}\right)}\right)}\right]\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}+\int_{0}^{2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\pi-\varphi}{2}\right)}\right)}\right]\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}-\int_{\pi}^{\pi-2\theta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}\right];~~~\small{\left[\varphi\mapsto\pi-\varphi\right]}\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\left[\operatorname{Cl}_{2}{\left(2\theta\right)}+\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}-\operatorname{Cl}_{2}{\left(\pi\right)}\right]\\ &=\theta\ln{\left(\tan{\left(\theta\right)}\right)}+\frac12\operatorname{Cl}_{2}{\left(2\theta\right)}+\frac12\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}.\\ \end{align}$$

We can use this relation to rewrite the expression for $\mathcal{I}$ obtained above as

$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(a+\sin{\left(\varphi\right)}\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi+\alpha\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}+\operatorname{Cl}_{2}{\left(-\pi-\alpha\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\alpha\right)}+\operatorname{Cl}_{2}{\left(\pi-\alpha\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}-\alpha\ln{\left(\tan{\left(\frac{\alpha}{2}\right)}\right)}+2\operatorname{Ti}_{2}{\left(\tan{\left(\frac{\alpha}{2}\right)}\right)}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}-\arcsin{\left(a\right)}\ln{\left(\frac{a}{1+\sqrt{1-a^{2}}}\right)}+2\operatorname{Ti}_{2}{\left(\frac{a}{1+\sqrt{1-a^{2}}}\right)}.\blacksquare\\ \end{align}$$

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