Solution of given definite integral: $\int _0^a \frac1x \tan ^{-1} \left(\frac{x^2}{d^2+x^2} \right)dx$

Question

Can following integral be solved in terms of elementary functions?

$$\int _{0}^{a} \frac{1}{x} \tan ^{-1} \bigg(\frac{x^2}{d^2+x^2} \bigg)dx$$

where $a,d \in R$

Could someone help me with this?

Answer 1

Could someone help me with this ?

Yes. I can. Here are two facts for future reference :

• the presence of $\dfrac{{\bf d}x}x$ usually begs for a linear substitution, since, for $x=\alpha~t,$ we have $$\dfrac{{\bf d}x}x=\dfrac{{\bf d}(\alpha~t)}{\alpha~t}=\dfrac{\alpha~{\bf d}t}{\alpha~t}=\dfrac{{\bf d}t}t.$$

• the presence of $1+x^2$ usually requires a trigonometric or hyperbolic substitution,
  since, for $x=\tan(t),$ we have $1+x^2=1+\tan^2t=\sec^2t,$ and, for $x=\sinh(u),$ we can
  write $1+x^2=1+\sinh^2u$ $=\cosh^2u.$

All in all, it would follow that we should employ the substitution $x=d\tan(t).$ Then, rewriting

$\sin^2t=\dfrac{1-\cos(2t)}2$ and $\sec(t)\csc(t)=\dfrac1{\sin(t)\cos(t)}=\dfrac2{\sin(2t)}=\dfrac1{\sqrt{1-\cos^2(2t)}},$ it only

feels natural to let $u=\cos(2t).$ Of course, this could all have been handled much more directly

by rewriting $\dfrac1x=\dfrac x{x^2}=\dfrac12\cdot\dfrac{{\bf d}(x^2)}{x^2},$ and then simply substituting $v=x^2,$ but where would

be the fun in that, right ? ;-$)$ Either way, the last and final steps involve yet another linear

substitution, $w=\dfrac{1-u}2,$ followed by factoring the integrand's new denominator using the

famous formula $A^2-B^2=(A-B)(A+B),$ and then employing some very basic partial

fraction decomposition, only to be left with a linear combination of integrals of the form

$\displaystyle\int\dfrac{\arctan(x)}x~dx$ and $\displaystyle\int\dfrac{\arctan(x)}{1-x}~dx,$ none of which are even remotely elementary.

Answer 2

Incomplete solution:

$a>0$.

$\displaystyle J(a)=\int_0^a \arctan\left(\dfrac{x^2}{d^2+x^2}\right)\cdot \dfrac{1}{x}dx$

Perform the change of variable $y=\dfrac{x^2}{d^2+x^2}$

$\begin{align}\displaystyle J(a)&=\int_0^{\tfrac{a^2}{d^2+a^2}} \dfrac{\arctan x}{2x(1-x)}dx\\ &=\dfrac{1}{2}\left(\int_0^{\tfrac{a^2}{d^2+a^2}} \dfrac{\arctan x}{x}dx+\int_0^{\tfrac{a^2}{d^2+a^2}} \dfrac{\arctan x}{1-x}dx\right) \end{align}$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{a}{1 \over x}\arctan\pars{x^2 \over d^{2} + x^{2}}\,\dd x = \int_{0}^{a/d}{1 \over x}\arctan\pars{x^2 \over x^{2} + 1}\,\dd x \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=} &\,\,\, {1 \over 2}\int_{0}^{\pars{a/d}^{\,\,2}}{1 \over x} \arctan\pars{x \over x + 1}\,\dd x\quad\pars{~next, integrate\ by\ parts~} \\[5mm] = &\ {1 \over 2}\,\ln\pars{\bracks{a \over d}^{2}} \arctan\pars{\bracks{a/d}^{2} \over \bracks{a/d}^{2} + 1} - {1 \over 4}\int_{0}^{\pars{a/d}^{\,\,2}}{\ln\pars{x} \over x^{2} + x + 1/2} \,\dd x \\[5mm] = &\ \ln\pars{\verts{a \over d}} \arctan\pars{a^{2} \over a^{2} + d^{2}}\ -\ {1 \over 4}\ \underbrace{\int_{0}^{\pars{a/d}^{\,\,2}} {\ln\pars{x} \over \pars{x - r}\pars{x - \bar{r}}}\,\dd x} _{\ds{\mc{J}}}\label{1}\tag{1} \end{align} where $\ds{r \equiv -\,{1 \over 2} - {1 \over 2}\,\ic}$.

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Then, \begin{align} \mc{J} & = \int_{0}^{\pars{a/d}^{\,\,2}} {\ln\pars{x} \over \pars{x - r}\pars{x - \bar{r}}}\,\dd x = \int_{0}^{\pars{a/d}^{\,\,2}}\ln\pars{x} \pars{{1 \over r - x} - {1 \over \bar{r} - x}}\,{1 \over \bar{r} - r}\,\dd x \\[5mm] & = 2\,\Im\int_{0}^{\pars{a/d}^{\,\,2}}{\ln\pars{x} \over r - x}\,\dd x = 2\,\Im\int_{0}^{2\pars{a/d}^{\,\,2}\,\bar{r}}{\ln\pars{rx} \over 1 - x}\,\dd x \\[5mm] & = -2\,\Im\pars{\ln\pars{1 - 2\bracks{a \over d}^{2}\bar{r}}\ln\pars{2\bracks{a \over d}^{2}r\bar{r}}} + 2\,\Im\int_{0}^{2\pars{a/d}^{\,\,2}\,\bar{r}}\ \overbrace{{\ln\pars{1 - x} \over x}}^{\ds{-\mrm{Li}_{2}\,'\pars{x}}}\,\dd x \\[5mm] & = -2\ln\pars{\verts{a \over d}}\,\Im\ln\pars{1 + \bracks{a \over d}^{2} - \bracks{a \over d}^{2}\ic} - 2\,\Im\mrm{Li}_{2}\pars{2\bracks{a \over d}^{2}\bar{r}} \\[5mm] & = 2\ln\pars{\verts{a \over d}}\,\arctan\pars{a^{2} \over a^{2} + d^{2}} - 2\,\Im\mrm{Li}_{2}\pars{-\bracks{a \over d}^{2} + \bracks{a \over d}^{2}\ic} = \bbx{\ds{\mc{J}}} \end{align}

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With expression \eqref{1}: \begin{align} &\int_{0}^{a}{1 \over x}\arctan\pars{x^2 \over d^{2} + x^{2}}\,\dd x \\[5mm] = &\ \bbx{\ds{% {1 \over 2}\ln\pars{\verts{a \over d}}\,\arctan\pars{a^{2} \over a^{2} + d^{2}} + {1 \over 2}\, \Im\mrm{Li}_{2}\pars{-\bracks{a \over d}^{2}\bracks{1 - \ic}}}} \end{align}

Answer 4

here is the indefinite integral by Maple $$\ln \left( x \right) \arctan \left( {\frac {{x}^{2}}{{d}^{2}+{x}^{2}} } \right) -1/2\,{d}^{2}\sum _{{\it \_R1}={\it RootOf} \left( 2\,{{\it \_Z}}^{4}+2\,{{\it \_Z}}^{2}{d}^{2}+{d}^{4} \right) }{\frac {1}{2\,{{ \it \_R1}}^{2}+{d}^{2}} \left( \ln \left( x \right) \ln \left( { \frac {{\it \_R1}-x}{{\it \_R1}}} \right) +{\it dilog} \left( {\frac { {\it \_R1}-x}{{\it \_R1}}} \right) \right) } $$