Any other methods for evaluating $\int_{0}^{\frac{\pi}{4}}\frac{\arctan(\cos(x))}{\cos(x)}\,dx$

Question

I evaluated the following integral:

$$I:=\int_{0}^{\frac{\pi}{4}}\frac{\arctan(\cos(x))}{\cos(x)}\,dx$$

I would like to see any alternate solutions, here is my work.

Using:

$$\frac{\arctan(x)}{x}\equiv\int_{0}^{1}\frac{1}{1+x^2y^2}\,dy$$

It is trivial to show that:

$$\frac{\arctan(\cos(x))}{\cos(x)}\equiv\int_{0}^{1}\frac{\sec^2(x)}{\tan^2(x)+1+y^2}\,dy$$

Therefore:

$$I=\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}\frac{\sec^2(x)}{\tan^2(x)+1+y^2}\,dy\,dx$$

Let

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

$$I=\int\int_{[0,1]^2}\frac{1}{1+x^2+y^2}\,dy\,dx$$

$$=\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sec(\theta)}\frac{r}{1+r^2}\,dr\,d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\csc(\theta)}\frac{r}{1+r^2}\,dr\,d\theta$$

$$=\frac{1}{2}\left[\int_{0}^{\frac{\pi}{4}}\log\left(2+\tan^2(\theta)\right)\,d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\log\left(2+\cot^2(\theta)\right)\,d\theta\right]$$

For the rightmost integral let

$$\theta\longrightarrow{\frac{\pi}{2}-\theta}$$

To get:

$$I=\int_{0}^{\frac{\pi}{4}}\log\left(2+\tan^2(\theta)\right)\,d\theta$$

Now let

$$\tan(\theta)=x$$

Which yields:

$$I=\int_{0}^{1}\frac{\log\left(2+x^2\right)}{1+x^2}\,dx$$

Define $I(t)\forall t\in[0,1]$ to be:

$$I(t):=\int_{0}^{1}\frac{\log\left(1+t(1+x^2)\right)}{1+x^2}\,dx$$

We can see that:

$$I(0)=0$$

$$I(1)=I$$

And so it follows from the F.T.C. That:

$$I=\int_{0}^{1}I’(t)\,dt$$

And so to follows that:

$$I’(t)=\int_{0}^{1}\frac{1}{tx^2+1+t}\,dx$$

$$=\frac{\arctan\left(\sqrt{\frac{t}{1+t}}\right)}{\sqrt{t(1+t)}}$$

And so we have that:

$$I=\int_{0}^{1}\frac{\arctan\left(\sqrt{\frac{t}{1+t}}\right)}{\sqrt{t(1+t)}}$$

Let

$$\sqrt{\frac{t}{1+t}}=x$$

To get:

$$I=2\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan(x)}{1-x^2}\,dx$$

Let

$$x\longrightarrow{\frac{1-x}{1+x}}$$

$$I=\int_{3-2\sqrt{2}}^{1}\frac{\frac{\pi}{4}-\arctan(x)}{x}\,dx$$

$$=\frac{\pi}{4}\int_{3-2\sqrt{2}}^{1}\frac{1}{x}\,dx+\int_{0}^{3-2\sqrt{2}}\frac{\arctan(x)}{x}\,dx-\int_{0}^{1}\frac{\arctan(x)}{x}\,dx$$

$$=\frac{\pi}{2}\log\left(1+\sqrt{2}\right)+Ti_{2}\left(3-2\sqrt{2}\right)-G$$

And so we have the result:

$$\int_{0}^{\frac{\pi}{4}}\frac{\arctan(\cos(x))}{\cos(x)}\,dx=\frac{\pi}{2}\log\left(1+\sqrt{2}\right)+Ti_{2}\left(3-2\sqrt{2}\right)-G$$

Where:

$$G:=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$$

Is Catalan’s constant.

And

$$Ti_{2}(x):=\int_{0}^{x}\frac{\arctan(t)}{t}\,dt$$

Is the inverse tangent integral function.

As I mentioned in the beginning of my post. Is there any other unique methods for the evaluation of this integral?

Answer 1

Using the tangent hal-angle substitution, there is an antiderivative.

At a point, you properly wrote that $$I=\int_{0}^{1}\frac{\log\left(2+x^2\right)}{1+x^2}\,dx$$

Write $$\log(2+x^2)=\log(x+i \sqrt 2)+\log(x-i\sqrt 2)$$ $$\frac{1}{1+x^2}=\frac{1}{(x+i)(x-i)}=\frac i 2\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$

to face four integrals $$I(a,b)=\int_0 ^1 \frac{\log(x+i\,a)}{x+i\,b}\,dx$$ which are symple (if you accept polylogarithms).

Recombining all pieces $$I=\frac{1}{8} \pi \log \left(102+72 \sqrt{2}\right)-\frac{1}{2} \log (2) \cot ^{-1}\left(\sqrt{2}\right)-\tan ^{-1}\left(\sqrt{2}\right) \sinh ^{-1}(1)-$$ $$\frac{i}{2} (\text{Li}_2((1-i) \left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right) )-\text{Li}_2((1+i) \left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right) ))-$$ $$-\frac{i}{2} \left(\text{Li}_2\left((2+i)-2 (-1)^{1/4}\right)-\text{Li}_2\left((2-i)+2 (-1)^{3/4}\right)\right)$$ which is ugly compared to your result.

Answer 2

At a time, I hoped to do something with $$I=\int_{0}^{\frac{\pi}{4}}\frac{\arctan(\cos(x))}{\cos(x)}\,dx=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\int_{0}^{\frac{\pi}{4}} \cos ^{2 n}(x)\,dx $$ since

$$J_n=\int_{0}^{\frac{\pi}{4}} \cos ^{2 n}(x)\,dx =\frac{1}{2} \left(\frac{\sqrt{\pi } \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}-B_{\frac{1}{2}}\left(n+\frac{1}{2},\frac{1}{2}\right) \right)$$ which leads to $$I=\frac{\pi}{2} \sinh ^{-1}(1)-\frac 12 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\,\,B_{\frac{1}{2}}\left(n+\frac{1}{2},\frac{1}{2}\right)$$

@Mariusz Iwaniuk provided a simple expression for the second summation (have a look here).

So, as a total

$$\large\color{blue}{I=\frac{\pi}{2} \sinh ^{-1}(1)+\frac{1}{6} \, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};\frac{1}{9}\right) -C}$$

Answer 3

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Let $\mathcal{I}$ denote the value of the following definite integral:

$$\mathcal{I}:=\int_{0}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\frac{\arctan{\left(\cos{\left(\varphi\right)}\right)}}{\cos{\left(\varphi\right)}}.$$

I'd like to highlight a more direct route to the closed form expression for $\mathcal{I}$ in terms of the inverse tangent integral by exploiting an inverse trigonometric identity which we'll derive.

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$$\begin{align} \mathcal{I} &=\int_{0}^{\frac{\pi}{4}}\mathrm{d}\varphi\,\frac{\arctan{\left(\cos{\left(\varphi\right)}\right)}}{\cos{\left(\varphi\right)}}\\ &=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\cdot\frac{\arctan{\left(\sqrt{1-x^{2}}\right)}}{\sqrt{1-x^{2}}};~~~\small{\left[\varphi=\arcsin{\left(x\right)}\right]}\\ &=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arctan{\left(\sqrt{1-x^{2}}\right)}}{1-x^{2}}\\ &=\int_{1}^{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{\arctan{\left(\sqrt{\frac{4y}{\left(1+y\right)^{2}}}\right)}}{\left[\frac{4y}{\left(1+y\right)^{2}}\right]};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\int_{\left(\sqrt{2}-1\right)^{2}}^{1}\mathrm{d}y\,\frac{\arctan{\left(\frac{2\sqrt{y}}{1+y}\right)}}{2y}\\ &=\int_{\sqrt{2}-1}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{2t}{1+t^{2}}\right)}}{t};~~~\small{\left[\sqrt{y}=t\right]}.\\ \end{align}$$

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Recall the following formula for the difference of two arctangents:

$$\arctan{\left(x\right)}-\arctan{\left(y\right)}=\arctan{\left(\frac{x-y}{1+xy}\right)};~~~\small{xy>-1}.$$

Then, setting $x=\frac{t}{p}\land y=pt$, we have

$$\arctan{\left(\frac{t}{p}\right)}-\arctan{\left(pt\right)}=\arctan{\left(\frac{(p^{-1}-p)t}{1+t^{2}}\right)};~~~\small{p\neq0}.$$

It can be shown that $p$ satisfies $p^{-1}-p=2\land p>0\iff p=\sqrt{2}-1$.

Hence, the arctangent functions obeys the following identity:

$$\arctan{\left(\frac{2t}{1+t^{2}}\right)}=\arctan{\left(\frac{t}{\sqrt{2}-1}\right)}-\arctan{\left((\sqrt{2}-1)t\right)};~~~\small{t\in\mathbb{R}}.$$

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Thus,

$$\begin{align} \mathcal{I} &=\int_{\sqrt{2}-1}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{2t}{1+t^{2}}\right)}}{t}\\ &=\int_{\sqrt{2}-1}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{t}{\sqrt{2}-1}\right)}-\arctan{\left((\sqrt{2}-1)t\right)}}{t}\\ &=\int_{a}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{t}{a}\right)}-\arctan{\left(at\right)}}{t};~~~\small{\left[a:=\sqrt{2}-1\in(0,1)\right]}\\ &=\int_{a}^{1}\mathrm{d}t\,\frac{\arctan{\left(\frac{t}{a}\right)}}{t}-\int_{a}^{1}\mathrm{d}t\,\frac{\arctan{\left(at\right)}}{t}\\ &=\int_{1}^{\frac{1}{a}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{u};~~~\small{\left[t=au\right]}\\ &~~~~~-\int_{a^{2}}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x};~~~\small{\left[t=a^{-1}x\right]}\\ &=\int_{a}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{1}{x}\right)}}{x};~~~\small{\left[u=x^{-1}\right]}\\ &~~~~~-\int_{a^{2}}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}\\ &=\int_{a}^{1}\mathrm{d}x\,\frac{\left[\frac{\pi}{2}-\arctan{\left(x\right)}\right]}{x}-\int_{a^{2}}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}\\ &=\frac{\pi}{2}\int_{a}^{1}\mathrm{d}x\,\frac{1}{x}-\int_{a}^{1}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}-\int_{a^{2}}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}\\ &=-\frac{\pi}{2}\ln{\left(a\right)}-\int_{a^{2}}^{1}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{x}\\ &=\frac{\pi}{2}\ln{\left(\frac{1}{a}\right)}-\operatorname{Ti}_{2}{\left(1\right)}+\operatorname{Ti}_{2}{\left(a^{2}\right)}\\ &=\frac{\pi}{2}\ln{\left(1+\sqrt{2}\right)}-G+\operatorname{Ti}_{2}{\left(3-2\sqrt{2}\right)}.\blacksquare\\ \end{align}$$

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