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Question

Compute

\int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x

$\int_{0}^{\infty} \frac{\arctan(ax)}{x(1+x^{2})}dx$

The answer is:

$\frac{\pi}{2}\ln(1+a)$ when $a \geq 0$

$-\frac{\pi}{2}\ln(1-a)$ when $a < 0$

Here is my problem, and I can’t even dive into a appropriate first step. The most difficult part is the $\arctan$ which I have no idea to eliminate. Thank you for help.

Fabian · Accepted Answer · 2014-06-30 09:21:59Z

You want to evaluate

I (a) = \int \infty 0 d x arctan ( a x ) x ( 1 + x 2 ) .

$I(a) = \int_0^\infty\! dx\,\frac{\arctan(a x)}{x (1+x^2)}.$

It is easy to see (because $\arctan(0)=0$ ) that $I(0)=0$ . Moreover, we have

I' (a) = \int \infty 0 d x 1 ( 1 + x 2 ) ( 1 + a 2 x 2 ) .

$I'(a)= \int_0^\infty \!dx \frac{1}{(1+x^2)(1+ a^2 x^2)}.$ The last integral can be integrated by standard methods (e.g. partial fraction expansion), and we obtain

I' (a) = a arctan ( a x ) - arctan ( x ) a 2 - 1 ∣ ∣ ∣ ∣ \infty x = 0 = π 2 | a | - 1 a 2 - 1 = π 2 1 1 + | a | .

$I'(a) = \frac{a \arctan(a x) -\arctan(x)}{a^2 -1}\Bigg|_{x=0}^\infty = \frac{\pi}2 \frac{|a| -1}{a^2-1} = \frac{\pi}2 \frac{1}{1 +|a|}.$

Another integration yields the final result

I (a) = \int a 0 d b I' (b) = π 2 \int a 0 d b (1 + | b |) - 1 = π 2 sgn (a) ln (1 + | a |) .

$I(a)= \int_0^a\!db\,I'(b) = \frac{\pi}2 \int_0^a\!db\,(1+|b|)^{-1} = \frac\pi2 \mathop{\rm sgn}(a)\ln(1+|a|).$

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Random Variable · Answer 2 · 2014-06-30 15:25:12Z

up vote 6 down vote

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Another approach is to consider $\displaystyle f(z) = \frac{\ln(1-iaz)}{z(1+z^{2})}$ (where $a>0$ ) and integrate around a contour on the complex plane that consists of the real axis and the upper half of $|z|=R$ .

As $R \to \infty$ , $\displaystyle \int f(z) \ dz$ vanishes along the upper half of $|z|=R$ .

Therefore,

\int \infty - \infty ln | 1 - i a x | - i arctan ( a x ) x ( 1 + x 2 ) d x a m p; = \int \infty - \infty ln ( 1 + a 2 x 2 ) - i arctan ( a x ) x ( 1 + x 2 ) d x a m p; = 2 π i Res [f (z), i] a m p; = 2 π i lim z \to i ln ( 1 - i a z ) z ( z + i ) a m p; = - i π ln (1 + a) .

$\begin{align} \int_{-\infty}^{\infty} \frac{\ln |1-iax| - i \arctan (ax) }{x(1+x^{2})} \ dx &= \int_{-\infty}^{\infty} \frac{\ln(1+a^{2}x^{2}) - i \arctan(ax)}{x(1+x^{2})} \ dx \\ &= 2 \pi i \ \text{Res}[f(z),i] \\ &= 2 \pi i \ \lim_{z \to i} \frac{\ln(1-iaz)}{z(z+i)} \\ &= - i \pi \ln(1+a) . \end{align}$

But

\int \infty - \infty ln ( 1 + a 2 x 2 ) - i arctan ( a x ) x ( 1 + x 2 ) d x a m p; = \int \infty - \infty ln ( 1 + a 2 x 2 ) x ( 1 + x 2 ) d x - i \int \infty - \infty arctan ( a x ) x ( 1 + x 2 ) d x a m p; = 0 - 2 i \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x a m p; = - 2 i \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x

$\begin{align} \int_{-\infty}^{\infty} \frac{\ln (1+a^{2}x^{2}) - i \arctan (ax) }{x(1+x^{2})} \ dx &= \int_{-\infty}^{\infty} \frac{\ln(1+a^{2}x^{2})}{x(1+x^{2})} \ dx - i \int_{-\infty}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx \\ &= 0 - 2i \int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx \\ &= -2i \int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx \end{align}$

since the first integrand is odd and the second integrand is even.

So we have

- 2 i \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = - i π ln (1 + a)

$-2i \int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx = - i \pi \ln(1+a)$

which implies

\int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx = \frac{\pi}{2} \ln(1+a) \ , \ a&gt;0.

$\int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx = \frac{\pi}{2} \ln(1+a) \ , \ a>0.$

But since $\arctan$ is an odd function,

\int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = sgn (a) π 2 ln (1 + | a |) .

$\int_{0}^{\infty} \frac{\arctan (ax)}{x(1+x^{2})} \ dx = \text{sgn}(a) \frac{\pi}{2} \ln(1+|a|) .$

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edited Jun 30 at 15:25

answered Jun 30 at 11:33

Random Variable
11.1k11552

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            <span class="comment-copy">Look super awesome ... Thank you anyway :)</span>
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            <span class="comment-copy">This is a good demonstration of the usefulness of contour integration (+1). It is hard to beat a trick like differentiation under the integral, though :-)</span>
            &ndash;&nbsp;
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            <span class="comment-copy">@Sacheo  Thanks.</span>
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            <span class="comment-copy">@robjohn Thanks. The evaluation turned out to be surprisingly simple.</span>
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = 1 2 a ∣ ∣ a ∣ ∣ \int \infty - \infty arctan ( x ) x ( x 2 + a 2 ) d x a m p; = 1 2 a ∣ ∣ a ∣ ∣ \int \infty - \infty i 2 ln (1 - i x 1 + i x)          = arctan (x) 1 x ( x 2 + a 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int \infty - \infty ln ( 1 - i x ) x ( x 2 + a 2 ) d x

$\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =\half\,a\verts{a}\int_{-\infty}^{\infty} {\arctan\pars{x} \over x\pars{x^{2} + a^{2}}}\,\dd x \\[3mm]&=\half\,a\verts{a}\int_{-\infty}^{\infty}\ \overbrace{{\ic \over 2}\,\ln\pars{1 - \ic x \over 1 + \ic x}} ^{\ds{=\ \arctan\pars{x}}}\ {1\over x\pars{x^{2} + a^{2}}}\,\dd x =-\,\half\,a\verts{a}\,\Im\int_{-\infty}^{\infty} {\ln\pars{1 - \ic x} \over x\pars{x^{2} + a^{2}}}\,\dd x \end{align}$

Set $\ds{t \equiv 1 - \ic x\quad\imp\quad x = \pars{t - 1}\ic}$ :
$a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int 1 - \infty i 1 + \infty i ln ( t ) ( t - 1 ) i [ - ( t - 1 ) 2 + a 2 ] i d t a m p; = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int 1 + \infty i 1 - \infty i ln ( t ) ( t - 1 ) ( t - r - ) ( t - r + ) d t where r \pm = 1 \pm ∣ ∣ a ∣ ∣$ $\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =-\,\half\,a\verts{a}\,\Im\int_{1 + \infty\ic}^{1 - \infty\ic} {\ln\pars{t} \over \pars{t - 1}\ic\bracks{-\pars{t - 1}^{2} + a^{2}}}\,\ic\,\dd t \\[3mm]&=-\,\half\,a\verts{a}\,\Im\color{#00f}{% \int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{t}\over \pars{t - 1}\pars{t - r_{-}}\pars{t - r_{+}}}\,\dd t} \quad \mbox{where}\quad r_{\pm} = 1 \pm \verts{a} \end{align}$

Now, we’ll evaluate the $\ds{\color{#00f}{\mbox{'blue integral'}}}$ . We take the “ $\ds{\ln}$ -branch cut” along the negative $\ds{t}$ -semi-axis and close the contour in a semi-circle “to the right” $\ds{\pars{~t > 1~}}$ :

a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ [- 2 π i ln ( r + ) + i 0 ( r + - 1 ) ( r + - r - )] a m p; = π a ∣ ∣ a ∣ ∣ [ln ( 1 + ∣ ∣ a ∣ ∣ ) ∣ ∣ a ∣ ∣ ( 2 ∣ ∣ a ∣ ∣ )]

$\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =-\,\half\,a\verts{a}\,\Im\bracks{-2\pi\ic\, {\ln\pars{r_{+}} + \ic 0 \over \pars{r_{+} - 1}\pars{r_{+} - r_{-}}}} \\[3mm]&=\pi\,a\verts{a}\,\bracks{% {\ln\pars{1 + \verts{a}} \over \verts{a}\pars{2\verts{a}}}} \end{align}$

$\int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = π 2 s g n (a) ln (1 + ∣ ∣ a ∣ ∣)$ $\color{#66f}{\large\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x ={\pi \over 2}\,\sgn\pars{a}\ln\pars{1 + \verts{a}}}$

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edited 9 hours ago

answered yesterday

Felix Marin
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Felix Marin · Answer 3 · 2014-07-03 13:12:24Z

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = 1 2 a ∣ ∣ a ∣ ∣ \int \infty - \infty arctan ( x ) x ( x 2 + a 2 ) d x a m p; = 1 2 a ∣ ∣ a ∣ ∣ \int \infty - \infty i 2 ln (1 - i x 1 + i x)          = arctan (x) 1 x ( x 2 + a 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int \infty - \infty ln ( 1 - i x ) x ( x 2 + a 2 ) d x

$\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =\half\,a\verts{a}\int_{-\infty}^{\infty} {\arctan\pars{x} \over x\pars{x^{2} + a^{2}}}\,\dd x \\[3mm]&=\half\,a\verts{a}\int_{-\infty}^{\infty}\ \overbrace{{\ic \over 2}\,\ln\pars{1 - \ic x \over 1 + \ic x}} ^{\ds{=\ \arctan\pars{x}}}\ {1\over x\pars{x^{2} + a^{2}}}\,\dd x =-\,\half\,a\verts{a}\,\Im\int_{-\infty}^{\infty} {\ln\pars{1 - \ic x} \over x\pars{x^{2} + a^{2}}}\,\dd x \end{align}$

Set $\ds{t \equiv 1 - \ic x\quad\imp\quad x = \pars{t - 1}\ic}$ :
$a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int 1 - \infty i 1 + \infty i ln ( t ) ( t - 1 ) i [ - ( t - 1 ) 2 + a 2 ] i d t a m p; = - 1 2 a ∣ ∣ a ∣ ∣ ℑ \int 1 + \infty i 1 - \infty i ln ( t ) ( t - 1 ) ( t - r - ) ( t - r + ) d t where r \pm = 1 \pm ∣ ∣ a ∣ ∣$ $\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =-\,\half\,a\verts{a}\,\Im\int_{1 + \infty\ic}^{1 - \infty\ic} {\ln\pars{t} \over \pars{t - 1}\ic\bracks{-\pars{t - 1}^{2} + a^{2}}}\,\ic\,\dd t \\[3mm]&=-\,\half\,a\verts{a}\,\Im\color{#00f}{% \int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{t}\over \pars{t - 1}\pars{t - r_{-}}\pars{t - r_{+}}}\,\dd t} \quad \mbox{where}\quad r_{\pm} = 1 \pm \verts{a} \end{align}$

Now, we’ll evaluate the $\ds{\color{#00f}{\mbox{'blue integral'}}}$ . We take the “ $\ds{\ln}$ -branch cut” along the negative $\ds{t}$ -semi-axis and close the contour in a semi-circle “to the right” $\ds{\pars{~t > 1~}}$ :

a m p; \int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = - 1 2 a ∣ ∣ a ∣ ∣ ℑ [- 2 π i ln ( r + ) + i 0 ( r + - 1 ) ( r + - r - )] a m p; = π a ∣ ∣ a ∣ ∣ [ln ( 1 + ∣ ∣ a ∣ ∣ ) ∣ ∣ a ∣ ∣ ( 2 ∣ ∣ a ∣ ∣ )]

$\begin{align}&\color{#c00000}{\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x} =-\,\half\,a\verts{a}\,\Im\bracks{-2\pi\ic\, {\ln\pars{r_{+}} + \ic 0 \over \pars{r_{+} - 1}\pars{r_{+} - r_{-}}}} \\[3mm]&=\pi\,a\verts{a}\,\bracks{% {\ln\pars{1 + \verts{a}} \over \verts{a}\pars{2\verts{a}}}} \end{align}$

$\int \infty 0 arctan ( a x ) x ( 1 + x 2 ) d x = π 2 s g n (a) ln (1 + ∣ ∣ a ∣ ∣)$ $\color{#66f}{\large\int_{0}^{\infty} {\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x ={\pi \over 2}\,\sgn\pars{a}\ln\pars{1 + \verts{a}}}$

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Thursday, July 3, 2014

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