Thursday, July 3, 2014

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integration - compute one improper integral involving arctangent - Mathematics Stack Exchange
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            <h1 itemprop="name"><a href="/questions/852026/compute-one-improper-integral-involving-arctangent" class="question-hyperlink">compute one improper integral involving arctangent</a></h1>
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    Compute

    0arctan(ax)x(1+x2)dx

    The answer is:

    π2ln(1+a) when a0

    π2ln(1a) when a &lt; 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.




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                <span class="comment-copy">Differentiate w.r.t $a$ and you will get an easier integral.</span>
                &ndash;&nbsp;
                    <a href="/users/35472/mhenni-benghorbal"
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                       class="comment-user">Mhenni Benghorbal</a>
                <span class="comment-date" dir="ltr"><span title="2014-06-30 09:07:54Z" class="relativetime-clean">Jun 30 at 9:07</span></span>
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                <span class="comment-copy">The correct result is $ \mathop{\text sgn}(a) \pi \ln(1+|a|)/2$</span>
                &ndash;&nbsp;
                    <a href="/users/7266/fabian"
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                <span class="comment-date" dir="ltr"><span title="2014-06-30 09:10:46Z" class="relativetime-clean">Jun 30 at 9:10</span></span>
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                <span class="comment-copy">I have add the missing little &#39;minus&#39; on the second condition. sorry for that.</span>
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                    <a href="/users/149685/zhen-zhang"
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                <span class="comment-date" dir="ltr"><span title="2014-06-30 09:17:03Z" class="relativetime-clean">Jun 30 at 9:17</span></span>
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    up vote 10 down vote accepted
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    You want to evaluate

    I(a)=0dxarctan(ax)x(1+x2).

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

    I(a)=0dx1(1+x2)(1+a2x2).
    The last integral can be integrated by standard methods (e.g. partial fraction expansion), and we obtain
    I(a)=aarctan(ax)arctan(x)a21x=0=π2|a|1a21=π211+|a|.

    Another integration yields the final result

    I(a)=a0dbI(b)=π2a0db(1+|b|)1=π2sgn(a)ln(1+|a|).

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    Another approach is to consider f(z)=ln(1iaz)z(1+z2) (where a&gt;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, f(z) dz vanishes along the upper half of |z|=R.

    Therefore,

    ln|1iax|iarctan(ax)x(1+x2) dxamp;=ln(1+a2x2)iarctan(ax)x(1+x2) dxamp;=2πi Res[f(z),i]amp;=2πi limziln(1iaz)z(z+i)amp;=iπln(1+a).

    But

    ln(1+a2x2)iarctan(ax)x(1+x2) dxamp;=ln(1+a2x2)x(1+x2) dxiarctan(ax)x(1+x2) dxamp;=02i0arctan(ax)x(1+x2) dxamp;=2i0arctan(ax)x(1+x2) dx

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

    So we have

    2i0arctan(ax)x(1+x2) dx=iπln(1+a)

    which implies

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

    But since arctan is an odd function,

    0arctan(ax)x(1+x2) dx=sgn(a)π2ln(1+|a|).

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                <span class="comment-copy">Look super awesome ... Thank you anyway :)</span>
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                    <a href="/users/149685/zhen-zhang"
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                <span class="comment-date" dir="ltr"><span title="2014-06-30 11:36:03Z" class="relativetime-clean">Jun 30 at 11:36</span></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;
                    <a href="/users/13854/robjohn"
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                       class="comment-user">robjohn<span class="mod-flair" title="moderator">&#9830;</span></a>
                <span class="comment-date" dir="ltr"><span title="2014-06-30 17:42:42Z" class="relativetime-clean">Jun 30 at 17:42</span></span>
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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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                <span class="comment-date" dir="ltr"><span title="2014-06-30 20:06:50Z" class="relativetime-clean">Jun 30 at 20:06</span></span>
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    amp;0arctan(ax)x(1+x2)dx=12aaarctan(x)x(x2+a2)dxamp;=12aa i2ln(1ix1+ix)= arctan(x) 1x(x2+a2)dx=12aaln(1ix)x(x2+a2)dx

    Set t1ixx=(t1)i:

    amp;0arctan(ax)x(1+x2)dx=12aa1i1+iln(t)(t1)i[(t1)2+a2]idtamp;=12aa1+i1iln(t)(t1)(tr)(tr+)dtwherer±=1±a

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

    amp;0arctan(ax)x(1+x2)dx=12aa[2πiln(r+)+i0(r+1)(r+r)]amp;=πaa[ln(1+a)a(2a)]

    0arctan(ax)x(1+x2)dx=π2sgn(a)ln(1+a)

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