MHB What is the integral of the Arcus function?

[Theia]

The exact value of $$\int_{-1}^1 \arcsin (x) \arccos (x) \arctan(x) \mathrm{d} x.$$

 

[Theia]

Suggested solution

Spoiler

We have
$$ I = \int_{-1} ^{1} \arcsin (x) \arccos (x) \arctan (x) \mathrm{d}x. $$

This procedure is fairly long, so I'll put some subsections to keep you on track.

Getting rid of arccos(x)

To start, one can separate positive and negative intervals and substitute \( x = -t \) to the negative interval part. We obtain

$$\begin{align*}I &= \int_{-1} ^{0} \arcsin (x) \arccos (x) \arctan (x) \mathrm{d}x + \int_{0} ^{1} \arcsin (x) \arccos (x) \arctan (x) \mathrm{d}x \\ &= -\int_{1} ^0 \arcsin (-t) \arccos (-t) \arctan (-t) \mathrm{d}t + \int_{0} ^{1} \arcsin (t) \arccos (t) \arctan (t) \mathrm{d}t \\ &= \int_{0} ^{1} \left( \arcsin (x) \left(\pi - \arccos (x) \right) \arctan (x) + \arcsin (x) \arccos (x) \arctan (x) \right) \mathrm{d}x \\ &= \pi \int_{0} ^{1} \arcsin (x) \arctan (x) \mathrm{d}x. \end{align*}$$

Getting rid of arcsin(x) and arctan(x)

Since it's easier to integrate \(\sin\) than \(\arcsin\), we'll next do integration by parts followed by a substitution. Let's choose \(f = \arctan (x)\) and \(g' = \arcsin (x)\) and write

$$\begin{align*}I &= \pi \Bigg/_{\hspace{-3mm}0}^{\,1} \arctan (x) \left( x\arcsin (x) + \sqrt{1 - x^{2}} \right) - \pi \int_{0}^{1} \frac{x\arcsin (x) + \sqrt{1 - x^{2}}}{1+x^{2}} \mathrm{d}x \\ &= \pi \cdot \frac{\pi}{4} \cdot \frac{\pi}{2} - \pi \int_{0}^{1} \frac{x\arcsin (x) + \sqrt{1 - x^{2}}}{1+x^{2}} \mathrm{d}x . \end{align*}$$

Now, let's substitute \(x = \sin (t) .\)

$$\begin{align*} I &= \frac{\pi ^3}{8} - \pi \int_0 ^{\pi/2} \frac{t\sin (t) + \cos (t)}{1 + \sin ^2 (t)} \ \cos (t)\mathrm{d}t \\ &= \frac{\pi ^3}{8} - \pi \ II - \pi \ III, \end{align*}$$

with obvious definitions for \(II\) and \(III\):

$$\begin{align*}II &= \int_0 ^{\pi/2} \frac{t \sin (t) \cos (t)}{1 + \sin ^2 (t)} \ \mathrm{d}t \\ III &= \int_0 ^{\pi/2} \frac{\cos ^2 (t)}{1 + \sin ^2 (t)} \ \mathrm{d}t. \end{align*}$$

The integral \(III\) seems somewhat easy, so at this point our interest points to the integral \(II.\)

Integration of the integral II

Again we integrate by parts with choices \(f = t\) and \(g' = \frac{\sin (t) \cos (t)}{1 + \sin ^2 (t)}.\) Hence we obtain

$$\begin{align*}II &= \Bigg/_{\hspace{-3mm}0}^{\,\pi/2} \frac{t \ln \left( 1 + \sin ^2 (t) \right)}{2} - \frac{1}{2} \int_0 ^{\pi/2} \ln \left( 1 + \sin ^2 (t)\right)\mathrm{d}t \\ &= \frac{\pi \ln 2}{4} - \frac{1}{2} II_2, \end{align*}$$

where

$$II_2 = \int _{0} ^{\pi/2} \ln \left( 1 + \sin ^2 (t) \right) \mathrm{d}t.$$

This we can evaluate using differentiation under the integral sign. Let's define

$$F(a) = \int_{0} ^{\pi/2} \ln \left( 1 + a\sin ^2 (t) \right) \mathrm{d}t,$$

so that \(II_{2} = F(1).\) Next we differentiate \(F(a)\) with respect to \(a\) to obtain

$$\frac{\mathrm{d}F}{\mathrm{d}a} = \int _{0} ^{\pi/2} \frac{\sin ^2 (t)}{1 + a\sin ^2 (t)} \mathrm{d}t.$$

Now there are many ways to integrate this. One is to divide nominator and denominator by \(\sin ^2 (t)\) and substitute \(z = \cot (t).\) I'm not sure if this way offers any advantages over the use of direct Weierstrass substitution. Anyhow, I used it and obtained

$$\begin{align*}\frac{\mathrm{d}F}{\mathrm{d}a} &= \int _{0} ^{1} \frac{\left( \frac{2u}{1 + u^{2}} \right) ^{2}}{1 + a\left( \frac{2u}{1 + u^2} \right) ^2} \cdot \frac{2u\mathrm{d}u}{1 + u^2} \\ &= 8\int_{0} ^{1} \frac{u^2\mathrm{d}u}{(1 + u^2)(u^4 + 2bu^2 + 1)}, \end{align*}$$

where \(b = 1 + 2a.\) This is easy to integrate using partial fractions. A shorthand \(c = \sqrt{b^2 - 1}\) will be used in following.

$$\begin{align*}\frac{\mathrm{d}F}{\mathrm{d}a} &= \frac{2}{c(b-1)} \int_{0} ^{1} \left(\frac{2c}{u^2+1} + \frac{1-b-c}{u^2+b+c} + \frac{b-c-1}{u^2+b-c} \right)\mathrm{d}u \\ &= \frac{2}{c(b-1)} \left( \frac{\pi c}{2} + \frac{1-b-c}{\sqrt{b+c}} \arctan \left( \frac{1}{\sqrt{b+c}} \right) + \frac{b-c-1}{\sqrt{b-c}} \arctan \left( \frac{1}{\sqrt{b-c}} \right) \right) \\ &= \frac{\pi}{b-1} + \frac{2}{c(b-1)} \cdot \frac{1-b-c}{\sqrt{b+c}} \cdot \frac{\pi}{2} \\ &= \frac{\pi}{2a} \left( 1 - \frac{1}{\sqrt{a+1}} \right).\end{align*}$$

Hence we've obtained for the function \(F(a):\)

$$F(a) = \frac{\pi}{2} \int \left( \frac{1}{a} - \frac{1}{a\sqrt{a+1}}\right) \mathrm{d}a.$$

There are again many ways to integrate this. I did it this way:

$$\begin{align*}F(a) &= \frac{\pi \ln a}{2} - \frac{\pi}{2} \int \frac{\mathrm{d}a}{a\sqrt{a+1}} \\ &= \frac{\pi \ln a}{2} - \frac{\pi}{2} \int \left( \frac{\sqrt{a+1}}{a} - \frac{1}{\sqrt{a+1}} \right) \mathrm{d}a \\ &= \frac{\pi \ln a}{2} + \frac{\pi}{2} \ln \left( \frac{\sqrt{a+1} + 1}{\sqrt{a+1} - 1} \right) + C \\ &= \frac{\pi}{2} \ln \left( \frac{a(\sqrt{a+1} + 1)}{\sqrt{a+1} - 1} \right) + C.\end{align*}$$

To find out the value of the constant of integration, we calculate the function at point \(a = 0\) according to our definition

$$F(0) = \int_{0} ^{\pi/2} \ln (1) \mathrm{d}a = 0$$

and using the general result we derived

$$\begin{align*}F(0) &= \lim_{a\to 0+} \frac{\pi}{2} \ln \left( \frac{a(\sqrt{a+1} + 1)}{\sqrt{a+1} - 1} \right) + C \\ &= \frac{\pi}{2} \lim_{a \to 0+} \ln \left( \frac{a(1 + \frac{a}{2} - \frac{a^2}{8} + \cdots + 1)}{1 + \frac{a}{2} - \frac{a^2}{8}} + \cdots - 1 \right) + C \\ &= \frac{\pi}{2} \lim_{a\to 0+} \ln \left( \frac{2 + \frac{a}{2} + \cdots}{\frac{1}{2} - \frac{a}{8} + \cdots} \right) + C \\ &= \pi \ln 2 + C.\end{align*}$$

Since the value of the function must be the same, no matter which way is being used, we obtain \(C = -\pi \ln 2.\) Hence finally

$$F(a) = \frac{\pi}{2} \ln \left( \frac{a(\sqrt{a+1} + 1)}{\sqrt{a+1} - 1} \right) - \pi \ln 2,$$

and

$$\begin{align*}\int_{0} ^{\pi/2} \ln \left( 1 + \sin ^{2} (t) \right) \mathrm{d}t &= F(a) \\ &= \frac{\pi}{2} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right) - \pi \ln 2 \\ &= \pi \ln \left( \frac{\sqrt{2} + 1}{2} \right).\end{align*}$$

Yey! So we've left only the integral \(III.\)

Integration of integral III

Here we first use the double angle formulas to get rid of squares. After it, we substitute \(2t = u\) and apply Weierstrass substitution. In other words

$$\begin{align*}III &= \int_0 ^{\pi/2} \frac{\cos ^2 (t)}{1 + \sin ^2 (t)} \mathrm{d}t \\ &= \int_{0} ^{\pi/2} \frac{1 + \cos (2t)}{3 - \cos (2t)} \mathrm{d}t \\ &= \frac{1}{2} \int_{0} ^{\pi} \frac{1 + \cos (u)}{3 - \cos (u)} \mathrm{d}u \\ &= \frac{1}{2} \int_{0} ^{\infty} \frac{1 + \frac{1 - p^2}{1 + p^2}}{3 - \frac{1 - p^2}{1 + p^2}} \ \frac{2\mathrm{d}p}{1 + p^2} \\ &= \int_{0} ^{\infty} \frac{\mathrm{d}p}{(1 + p^2)(1 + 2p^2)} \\ &= \cdots \\ &= \frac{\pi (\sqrt{2} - 1)}{2}.\end{align*}$$

Final result

$$\begin{align*}I &= \frac{\pi ^3}{8} - \pi \ II - \pi \ III \\ &= \frac{\pi ^3}{8} - \pi \left(\frac{\pi \ln 2}{4} - \frac{1}{2} \pi \ln \left( \frac{\sqrt{2} + 1}{2} \right) \right) - \pi \cdot \frac{\pi (\sqrt{2} - 1)}{2} \\ &= \frac{\pi ^3}{8} - \frac{\pi ^2}{4} \ln \left( \frac{8}{3 + 2\sqrt{2}} \right) - \frac{\pi ^2 (\sqrt{2} - 1)}{2} \\ &\approx 1.05031\ldots\end{align*}$$