Evaluate $\displaystyle \int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$

My effort:

\begin{align*} I(a)&=\int^{2}_{0}\frac{\tan^{-1}(ax)}{1+4x}\mathrm dx\\ I'(a) &= \int^{2}_{0}\frac{x}{(1+4x)(1+a^2x^2)}\mathrm dx\\ I'(a) &= \frac{1}{4}\int^{2}_{0}\frac{(1+4x)-1}{(1+4x)(1+a^2x^2)}dx\\ I'(a) &= \frac{1}{4a}\tan^{-1}(2)-\frac{1}{4}\int^{2}_{0}\frac{1}{(1+4x)(1+a^2x^2)}dx \end{align*}

Then how to proceed? Thank you.

  • 4
    $\begingroup$ Partial Fraction Decomposition? $\endgroup$ – mrtaurho Mar 23 at 12:11
  • $\begingroup$ @ mrtaurho i searcing for some easy way to handle that problem but not find it.have any idea please help me thanks $\endgroup$ – jacky Mar 23 at 12:23
  • 1
    $\begingroup$ From where did you get this integral? Is it worth hoping an elementary answer? $\endgroup$ – Zacky Mar 23 at 13:13
  • 2
    $\begingroup$ What is $tan^{-1}$? Is it $arctan$ or $cot$? $\endgroup$ – Hume2 May 22 at 8:10
  • 1
    $\begingroup$ Thank you for awards! There were a creative command, and my proposition is to continue the competition in the new question - to express the integral $\int_0^p \dfrac{\arctan x}{1+qx}\,dx$ in the elementary functions. $\endgroup$ – Yuri Negometyanov May 30 at 18:45

Another way to obtain result.

Are known trigonometric equalities

$$\tan(a+b) = \dfrac{\tan a+\tan b}{1-\tan a\tan b},\tag{Q1}$$ $$\tan\left(a+\frac\pi4\right) = \dfrac{\tan a +1}{1-\tan a},\tag{Q2}$$ $$\sec^2\left(a+\frac\pi4\right) = 2\dfrac{1+\tan^2a}{(1-\tan a)^2}.\tag{Q3}$$

Let $\,x= \tan (y+\frac\pi4),\quad dx=\sec^2(y+\frac\pi4)\,dy,\,$ then

\begin{align} &I(1) = \int\limits_0^2 \dfrac{\frac\pi4+\arctan x-\frac\pi4}{1+4x}\,dx = \dfrac\pi4\int\limits_0^2 \dfrac{dx}{1+4x} + \int\limits_{\large-\frac\pi4}^{\large\arctan\frac13} \dfrac {y\sec^2(y+\frac\pi4)\,dy}{1+4\tan(y+\frac\pi4)}\\ & = \dfrac\pi{16}\ln(1+4x)\bigg|_0^2 + \int\limits_{\large-\frac\pi4}^{\large\arctan \frac13}\dfrac{2(1+\tan^2y)y\,dy} {(1-\tan y)^2\left(1+4\frac{\large\tan y+1}{\large1-\tan y}\right)}\\ & = \dfrac{\pi\ln3}8 + \int\limits_{\large-\frac\pi4}^{\large\arctan \frac13} \dfrac{2y\sec^2y\,dy}{(1-\tan y)(5+3\tan y)}\\ & = \dfrac{\pi\ln3}8 + \dfrac14\int\limits_{\large-\frac\pi4}^{\large\arctan \frac13} y\left(\dfrac1{1-\tan y}+\dfrac3{5+3\tan y}\right)\sec^2y\,dy\\ & = \dfrac{\pi\ln3}8 + \dfrac14\int\limits_{\large-\frac\pi4}^{\large\arctan \frac13} y\,d\ln\dfrac{5+3\tan y}{1-\tan y}\\ & = \dfrac{\pi\ln3}8 + \dfrac y4\,\ln\dfrac{5+3\tan y}{1-\tan y} \Bigg|_{\large-\frac\pi4}^{\large\arctan \frac13} - \dfrac14\int\limits_{\large-\frac\pi4}^{\large\arctan \frac13} \ln\dfrac{5+3\tan y}{1-\tan y}\,dy\\ & = \dfrac{\pi\ln3}8 + \dfrac{\dfrac\pi2-\arctan3}2\ \ln3 - \dfrac{\dfrac\pi2-\arctan3+\dfrac\pi4}4\ \ln5 - \dfrac14\int\limits_{\large-\frac\pi4}^{\large\arctan \frac13} \ln\dfrac{1+\dfrac35\tan y}{1-\tan y}\,dy,\\ \end{align} $$I(1) = \dfrac{3\pi - 4\arctan3}{16}\ \ln\dfrac95 - \dfrac14 J(p,y)\Bigg|_{\large y=-\frac\pi4}^{\large\arctan \frac13} \Bigg|_{\large p=-1}^{\large\frac35},\tag1$$

where $$J(p,y) = \int\,\ln(1+p\tan y)\,dy,\tag2$$ $$|p|\le 1,\quad |\tan y| \le1,\tag3$$

\begin{align} &J(p,y) = \dfrac i2 \left(\operatorname{Li}_2\left(\dfrac{1+p\tan y}{1-ip}\right) - \operatorname{Li}_2\left(\dfrac{1+p\tan y}{1+ip}\right)\right)\\ & + \dfrac i2\left(\ln\dfrac{1-i\tan y}{-1+ip}-\ln\dfrac{1+i\tan y}{-1-ip}\right) \ln(1+p\tan y) + \mathrm{constant} \end{align} (see also Wolfram Alpha),

$$\ \operatorname {Li}_2(z) = \sum\limits_{j=1}^\infty\dfrac{z^j}{j^2}\tag{Q4}$$

is the polylogarithm.

Is known that

$$\ln a = \ln|a| + i\arg a.\tag{Q5}$$

Taking in account conditions $(3),$ one can get \begin{align} &\dfrac i2\left(\ln\dfrac{1-i\tan y}{-1+ip} - \ln\dfrac{1+i\tan y}{-1-ip}\right) = \dfrac i2\left(\ln\dfrac{1-i\tan y}{1+i\tan y} + \ln\dfrac{1+ip}{1-ip}\right)\\ & = \dfrac i2\left(-2iy + 2i\arctan p\right) = y - \arctan p. \end{align}

Therefore, under the conditions $(3)$ can be used expression in the form of $$J(p,y) = \dfrac i2 \left(\operatorname{Li}_2\left(\dfrac{1+p\tan y}{1-ip}\right) - \operatorname{Li}_2\left(\dfrac{1+p\tan y}{1+ip}\right)\right) + \left(y-\arctan p\right)\ \ln(1+p\tan y).\tag4$$

From $(1),(4),(Q1)$ should $$I(1) = \dfrac{3\pi - 4\arctan3}{16}\ \ln\dfrac95 - \dfrac14 F(p,t)\Bigg|_{t=-1}\phantom{|\hspace{-38mu}}^{\large\frac13} \Bigg|_{p=-1}\phantom{|\hspace{-38mu}}^{\large\frac35},\tag5$$

where $$F(p,t) = \dfrac i2 \left(\operatorname{Li}_2\left(\dfrac{1+pt}{1-ip}\right) - \operatorname{Li}_2\left(\dfrac{1+pt}{1+ip}\right)\right) + \arctan\dfrac{t-p}{1+pt}\ \ln(1+pt).\tag6$$

Finally, $$\color{brown}{\boxed{\phantom{\bigg|\!}I_1\approx0.27442\,80145\,78530\ }}$$ (see also Wolfram Alpha calculations of constant and the final checking).

Numeric calculations give the same result.



Try$${x\over (1+4x)(1+a^2x^2)}={{-{4\over a^2+16}\over 1+4x}}+{{a^2x+4\over a^2+16}\over 1+a^2x^2}={1\over a^2+16}\left({{-{4}\over 1+4x}}+{{a^2x+4}\over 1+a^2x^2}\right)$$

  • $\begingroup$ Yep, but integrating $I'(a)$ back with respect to $a$ looks hopeless. $\endgroup$ – Alex M. May 22 at 16:45

When I tried to simplify result, I found more elegant way to compute this integral. We now that $\Im Log z= \phi=\arctan(\Im z/\Re z)$ Due to this I can rewrite the integral in more simple way $$\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx=\Im \int^{2}_{0}\frac{\log(1+ix)}{1+4x}\mathrm dx $$ It can be rewrite through logarith and dilogarithm by linear changing variable https://en.wikipedia.org/wiki/Spence%27s_function $$I=\frac{\Im}{4} \left(-\text{Li}_2\left(\frac{16}{17}+\frac{4 i}{17}\right)+\text{Li}_2\left(\frac{8}{17}+\frac{36 i}{17}\right)+\log \left(\frac{9}{17}-\frac{36 i}{17}\right) \log (1+2 i)\right) $$

I am not a specialist in the polylogarithm, probably the imaginary parts of polilogarithm can be simplified.

The result for with additional parameter $$ \frac{\Im}{4} \left(\text{Li}_2\left(\frac{4 i-8 a}{a+4 i}\right)-\text{Li}_2\left(\frac{4 i}{a+4 i}\right)+\log \left(\frac{9 a}{a+4 i}\right) \log (1+2 i a)\right) $$


Firstly, do the integration by parts and and decompose a denominator $$ \begin{align*} & \int_0^2 \frac{\tan^{-1} (a x)}{1+4x} dx \\ = &\ \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{4}\int_0^2 \frac{\log(1+4x)}{1+a^2x^2} dx\\ = &\ \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{8}\int_0^2 \log(1+4x)(\frac{1}{1+iax}+\frac{1}{1-iax}) dx \end{align*}$$ The obtained integrals can be rewrote through the dilogarithm $\text{Li}_2$ functions. The result have the following form

$$ \begin{align*} I(a)&= \dfrac{\log 3 \tan^{-1} (2a)}{2}+\frac{i}{8}\left(\text{Li}_2\left(\frac{a}{a-4 i}\right)-\text{Li}_2\left(\frac{9 a}{a-4 i}\right)-\text{Li}_2\left(\frac{a}{a+4 i}\right) \\ + \text{Li}_2\left(\frac{9 a}{a+4 i}\right)-\log (9) \log \left(\frac{8 a+4 i}{-a+4 i}\right)+ \log (9) \log \left(\frac{-8 a+4 i}{a+4 i}\right)\right) \end{align*} $$ I think this answer can be simplified, but I do not want to do it.

  • $\begingroup$ More suitable looks the common form of $\int_0^b \dfrac{\arctan x\,dx}{1+ax}$ $\endgroup$ – Yuri Negometyanov May 27 at 10:43

To calculate by series or to try to get a "closed form" ?

I think the best is to use here the following formula:

$$\int\frac{\ln(x+z)}{x-a}\,dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)~\text{Li}_1\left(\frac{x+z}{a+z}\right) + C$$

It’s $~\displaystyle\tan x = \Im\ln(1+ix) = \frac{\pi}{2} + \Im\ln(x-i)~$ and we can set $~\displaystyle(a,z):=(-\frac{1}{4},-i)~$ .

Then we get:

$\displaystyle \int\limits_0^2\frac{\arctan x}{1+4x}\,dx = \frac{\pi}{2}\int\limits_0^2\frac{dx}{1+4x} + \frac{1}{4}\Im\int\limits_0^2\frac{\ln(x-i)}{x-\left(-\frac{1}{4}\right)}\,dx$

$\displaystyle =\frac{\pi}{4}\ln 3 + \frac{1}{4}\Im\left(\text{Li}_2\left(\frac{8+i36}{17}\right) - \ln(2-i)~\text{Li}_1\left(\frac{8+i36}{17}\right) \right) $

$\displaystyle\hspace{1.8cm} - \frac{1}{4}\Im\left( \text{Li}_2\left(\frac{16+i4}{17}\right) - \ln(-i)~\text{Li}_1\left(\frac{16+i4}{17}\right) \right) $

Separation of real part and imaginary part:

$\displaystyle \ln(2-i) = \frac{\ln 5}{2} - i\arctan\frac{1}{2} \enspace , \enspace\enspace \ln(-i) = -i\frac{\pi}{2}$

$\displaystyle \text{Li}_1\left(\frac{8+i36}{17}\right) = -\frac{1}{2}\ln\frac{81}{17} + i\arctan 4 \enspace , \enspace\enspace \text{Li}_1\left(\frac{16+i4}{17}\right) = \frac{1}{2}\ln 17 + i\arctan 4$

Note: $\enspace$ For calculations with $~x>0~$ we can use $\displaystyle ~~\arctan x = \frac{\pi}{2} - \arctan\frac{1}{x}~$ .

$\displaystyle A:=\Im\text{Li}_2\left(\frac{16+i4}{17}\right) = \Im\text{Li}_2\left(\frac{\sqrt{272}}{17}e^{i\arctan\frac{1}{4}}\right)=\sum\limits_{k=1}^\infty\frac{\sin\left(k\arctan\frac{1}{4}\right)}{k^2}\left(\frac{\sqrt{272}}{17}\right)^k$

Using $~\displaystyle \text{Li}_2(z) = -\text{Li}_2\left(\frac{1}{z}\right) -\frac{\pi^2}{6}-\frac{1}{2}\ln^2(-z)~$ we get:

$\displaystyle \Im\text{Li}_2\left(\frac{8+i36}{17}\right) = \Im\text{Li}_2\left(\frac{\sqrt{1360}}{17}e^{i\arctan\frac{9}{2}}\right) = B - \left(\ln\frac{\sqrt{1360}}{17}\right)\left(-\pi + \arctan\frac{9}{2}\right)$

with $\enspace \displaystyle B := \sum\limits_{k=1}^\infty\frac{\sin\left(k\arctan\frac{9}{2}\right)}{k^2}\left(\frac{17}{\sqrt{1360}}\right)^k$

$\displaystyle R := \pi\ln\frac{36}{17} + \left(\ln\frac{85}{81}\right)\left(\arctan\frac{1}{4}\right) + \left(\ln\frac{80}{81}\right)\left(\arctan\frac{2}{9}\right)$

$A\approx 0.54772569304105608037669001217295320069571309$

$B\approx 0.46231444614312936610749685795341278946044980$

$R\approx 2.36624661042409606268346795278249163083184498$

$\displaystyle \int\limits_0^2\frac{\arctan x}{1+4x}\,dx = \frac{R}{8} + \frac{B-A}{4} \approx 0.2744280145785303292681352055429263510451648$


$\def\B{\mathrm{B}}$Here we find the value of the integral as an explicitly real sum with good convergence properties, equation (1) below.

Let $x=1+t$ so $$I = \frac{1}{5} \int_{-1}^1 \frac{\arctan(1+t)}{1+\frac{4}{5}t}dt.$$ But $$\frac{1}{1+\frac 4 5 t} = \sum_{k=0}^\infty a_k t^k \quad\textrm{and}\quad \arctan(1+t) = \sum_{k=0}^\infty b_k t^k,$$ where $$a_k = (-4/5)^k,\, b_0 = \frac \pi 4, \textrm{ and } b_k = (-1)^{k+1}\frac{\sin(k\pi/4)}{k2^{k/2}} \textrm{ for }k\geq 1.$$ Thus, \begin{align*} I &= \frac{1}{5} \int_{-1}^1 \sum_{k=0}^\infty c_k t^k dt = \frac{2}{5}\sum_{k=0}^\infty \frac{c_{2k}}{2k+1}, \end{align*} where $$c_k = \sum_{j=0}^k a_{k-j} b_j$$ is the Cauchy product of $a_k$ and $b_k$. This can be rewritten as \begin{align*} I &= \frac \pi 8 \log 3 - \frac 2 5 \sum_{k=1}^\infty \sum_{j=1}^{2k} \frac 1 {2k+1} \left(\frac 4 5\right)^{2k-j} \frac{\sin(j\pi/4)}{j 2^{j/2}} \\ &= \frac \pi 8 \log 3 - \frac 2 5 \sum_{j=1}^\infty \sum_{k=\lceil j/2\rceil}^{\infty} \frac 1 {2k+1} \left(\frac 4 5\right)^{2k-j} \frac{\sin(j\pi/4)}{j 2^{j/2}} \\ &= \frac \pi 8 \log 3 - \frac 1 4 \sum_{j=1}^\infty \underbrace{\left(\frac 5 4\right)^{j} \frac{\sin(j\pi/4)}{j 2^{j/2}} \B({16}/{25};{1}/{2}+\lceil{j}/{2}\rceil,0)}_{d_j}, \end{align*} where $$\B(x;a,b) = \int_0^x t^{a-1}(1-t)^{b-1}dt$$ is the incomplete beta function. We then write the sum in the following way, $$\sum_{j=1}^\infty d_j = \sum_{n=0}^\infty (d_{4n+1}+d_{4n+2}+d_{4n+3}+d_{4n+4}).$$ Note that $d_{4n+4}=0$. By exploiting the properties of the incomplete beta function $d_{4n+1}+d_{4n+2}+d_{4n+3}$ may be combined. In particular we use that $$\B(x;a+1,b) = \B(x;a,b) - \frac{x^a(1-x)^b}{a}.$$ After some manipulation we find \begin{align*} I &= \frac \pi 8 \log 3 + \frac{1}{128}\Phi(-1/4,2,3/4) \\ & \quad - \frac{5}{1024}\sum_{n=0}^\infty (-1)^n \left(\frac{25}{32}\right)^{2n} \frac{776n^2+790n+181}{(2n+1)(4n+1)(4n+3)} \\ &\hspace{15ex}\times\B(16/25;2n+3/2,0),\tag{1} \end{align*} where $\Phi(z,s,\alpha) = \sum_{n=0}^\infty {z^n}/{(n+\alpha)^s}$ is the Lerch transcendant. (Note that $\Phi(-1/4,2,3/4)\approx 1.703446578$.) We define $I_N$ to be given by (1) with the replacement $\sum_{n=0}^\infty\rightarrow\sum_{n=0}^N$ and show some of the partial sums to ten decimal places below. $$\begin{array}{l|l} N & I_N \\ \hline 0 & 0.2687920302 \\ 1 & 0.2749812408 \\ 2 & 0.2743541958 \\ 3 & 0.2744394932 \\ 4 & 0.2744260568 \\ 5 & 0.2744283699 \\ 6 & 0.2744279472 \\ 7 & 0.2744280278 \\ 8 & 0.2744280119 \\ 9 & 0.2744280151 \\ 10 & 0.2744280145 \\ 11 & 0.2744280146 \\ 12 & 0.2744280146 \\ \end{array}$$

  • $\begingroup$ Wolfram Alpha can not calculate this sum in the closed form. Besides, this series diverges from $n=80.$ $\endgroup$ – Yuri Negometyanov May 28 at 21:35
  • $\begingroup$ @YuriNegometyanov: I appreciate the close reading. See the claim at the top of the post. The series does not diverge. It is straightforward to show that $\mathrm{B}(16/25;2n+3/2,0)\le (2^7/5)(4/5)^{4n}/(36n+27)$. $\endgroup$ – user26872 May 28 at 22:05

Yes,this integral seems hopeless from the point of view that we could get some simple closed form solution.

As a practical person who has to solve problems in real life where it is necessary to calculate not with great precision I use often the following approach (applied to given issue):

I'm trying to replace the integrand in the integral with a simpler expression so that the latter would differ as little as possible from the value of the original expression.

In this case I will replace $\arctan x$ with

$$\frac{\arctan 2}{26}x(23-5x)$$

The maximum deviation of this expression from $\arctan x$ in $[0,2]$ is less than $0.03$

Now, replacing $\arctan x$ in the integral with the expression I evaluate the integral and get

$$\frac{\arctan 2}{832}(308-97\ln 3)$$ The absolute error of this result is about $0.007$

This is even better result than expected.

Now, noting that $\arctan 2$ and $\ln 3$ are both very close to $1$ I simplify the result to $\frac{211}{832}$

The absolute error of this last is still satisfactory from a practical point of view (about $0.021$)

In similar manner it is possible quickly estimate many difficult integrals (often easier than numerical integration procedures)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.