MHB What is the integral of the Arcus function?

  • Thread starter Thread starter Theia
  • Start date Start date
  • Tags Tags
    Function Integral
Theia
Messages
121
Reaction score
1
The exact value of $$\int_{-1}^1 \arcsin (x) \arccos (x) \arctan(x) \mathrm{d} x.$$
 
Mathematics news on Phys.org
Suggested solution

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

$$\begin{align*}I &= \frac{\pi ^3}{8} - \pi \ II - \pi \ III \\ &= \frac{\pi ^3}{8} - \pi \left(\frac{\pi \ln 2}{4} - \frac{1}{2} \pi \ln \left( \frac{\sqrt{2} + 1}{2} \right) \right) - \pi \cdot \frac{\pi (\sqrt{2} - 1)}{2} \\ &= \frac{\pi ^3}{8} - \frac{\pi ^2}{4} \ln \left( \frac{8}{3 + 2\sqrt{2}} \right) - \frac{\pi ^2 (\sqrt{2} - 1)}{2} \\ &\approx 1.05031\ldots\end{align*}$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top