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How do we integrate forms of following type with imaginary terms involved? Can we get a closed form of it as result?

$\int-\frac{1}{4}(e^{-ix}-e^{ix})^2e^{\frac{4i(e^{-ix}-e^{ix})}{(e^{-ix}+e^{ix})^2}} \ dx \tag1$

Harry Peter
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Nirvana
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    take $e^{ix}=cosx + isinx$ and $e^{-ix}=cosx - isinx$ and then try to solve. – Jasser Sep 17 '14 at 10:38
  • Your recent question really make me wonder where you find them. – UserX Sep 17 '14 at 10:45
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    The integrand is Real. – Macavity Sep 17 '14 at 10:47
  • @UserX. Once I asked here if there a limit of the imagination of the textbooks, teachers, professors ... No answers, so I suppose that it is close to $\infty$ – Claude Leibovici Sep 17 '14 at 10:47
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    @Macavity only if you assume $\Im(x)=0$. Then the integrand is $\sin^2(x) e^{2 \tan(x) \sec(x)}$ – UserX Sep 17 '14 at 10:50
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    @UserX The integrand can be written as $\int\sin^2xe^{2\tan{x}\sec{x}}$ even if $x$ is complex. The relations hold in general. – David H Sep 17 '14 at 10:59
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    @DavidH In that case there is nevertheless no reason to assume the integrand to be real. But it seems natural to assume the integration to be over the real axis. – Jonas Dahlbæk Sep 17 '14 at 11:02
  • @DavidH Actually I came back this equation (1) from the trig form you mentioned .Because I couldn't find a proper form. Then I turned in to imaginary stuff..The trig form u mentioned cant be reduced to a closed form. In this group there are other approaches regarding imaginary integration here ..I am just thinking in that way.. Checking other stuffs like this too for getting some grip on complex integration – Nirvana Sep 17 '14 at 11:04
  • @UserX I came to this form from a trig function which I couldnot reduce to a proper form. I had mentioned in my previous reply . Thanks – Nirvana Sep 17 '14 at 11:08

1 Answers1

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$\int-\dfrac{1}{4}(e^{-ix}-e^{ix})^2e^\frac{4i(e^{-ix}-e^{ix})}{(e^{-ix}+e^{ix})^2}~dx$

$=\int(\sin^2x)e^{2\sec x\tan x}~dx$

$=\int(1-\cos^2x)\sum\limits_{n=0}^\infty\dfrac{4^n\sec^{2n}x\tan^{2n}x}{(2n)!}dx+\int(1-\cos^2x)\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n+1}x\tan^{2n+1}x}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{4^n\sec^{2n}x\tan^{2n}x}{(2n)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{4^n\sec^{2n-2}x\tan^{2n}x}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n+1}x\tan^{2n+1}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n-1}x\tan^{2n+1}x}{(2n+1)!}dx$

$=\int dx+\int\sum\limits_{n=1}^\infty\dfrac{4^n\sec^{2n}x\tan^{2n}x}{(2n)!}dx-\int\cos^2x~dx-2\int\tan^2x~dx-\int\sum\limits_{n=2}^\infty\dfrac{4^n\sec^{2n-2}x\tan^{2n}x}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n+1}x\tan^{2n+1}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n-1}x\tan^{2n+1}x}{(2n+1)!}dx$

$=\int\sin^2x~dx-2\int\tan^2x~dx+\int\sum\limits_{n=0}^\infty\dfrac{4^{n+1}\sec^{2n+2}x\tan^{2n+2}x}{(2n+2)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{4^{n+2}\sec^{2n+2}x\tan^{2n+4}x}{(2n+4)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n+1}x\tan^{2n+1}x}{(2n+1)!}dx-\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n-1}x\tan^{2n+1}x}{(2n+1)!}dx$

$=\int\dfrac{1-\cos2x}{2}dx-2\int(\sec^2x-1)~dx+\int\sum\limits_{n=0}^\infty\dfrac{4^{n+1}\sec^{2n}x\tan^{2n+2}x}{(2n+2)!}d(\tan x)-\int\sum\limits_{n=0}^\infty\dfrac{4^{n+2}\sec^{2n}x\tan^{2n+4}x}{(2n+4)!}d(\tan x)+\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n}x\tan^{2n}x}{(2n+1)!}d(\sec x)-\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\sec^{2n-2}x\tan^{2n}x}{(2n+1)!}d(\sec x)$

$=\int\dfrac{3}{2}dx-\int\dfrac{\cos2x}{2}dx-2\int\sec^2x~dx+\int\sum\limits_{n=0}^\infty\dfrac{4^{n+1}(1+\tan^2x)^n\tan^{2n+2}x}{(2n+2)!}d(\tan x)-\int\sum\limits_{n=0}^\infty\dfrac{4^{n+2}(1+\tan^2x)^n\tan^{2n+4}x}{(2n+4)!}d(\tan x)+\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}(\sec^2x-1)^n\sec^{2n}x}{(2n+1)!}d(\sec x)-\int\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}(\sec^2x-1)^n\sec^{2n-2}x}{(2n+1)!}d(\sec x)$

$=\int\dfrac{3}{2}dx-\int\dfrac{\cos2x}{2}dx-2\int\sec^2x~dx+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+1}C_k^n\tan^{2k}x\tan^{2n+2}x}{(2n+2)!}d(\tan x)-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+2}C_k^n\tan^{2k}x\tan^{2n+4}x}{(2n+4)!}d(\tan x)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2^{2n+1}C_k^n(-1)^{n-k}\sec^{2k}x\sec^{2n}x}{(2n+1)!}d(\sec x)-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2^{2n+1}C_k^n(-1)^{n-k}\sec^{2k}x\sec^{2n-2}x}{(2n+1)!}d(\sec x)$

$=\int\dfrac{3}{2}dx-\int\dfrac{\cos2x}{2}dx-2\int\sec^2x~dx+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+1}n!\tan^{2n+2k+2}x}{(2n+2)!k!(n-k)!}d(\tan x)-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+2}n!\tan^{2n+2k+4}x}{(2n+4)!k!(n-k)!}d(\tan x)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}2^{2n+1}\sec^{2n+2k}x}{(2n+1)!k!(n-k)!}d(\sec x)-\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}2^{2n+1}\sec^{2n+2k-2}x}{(2n+1)!k!(n-k)!}d(\sec x)$

$=\dfrac{3x}{2}-\dfrac{\sin2x}{4}-2\tan x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+1}n!\tan^{2n+2k+3}x}{(2n+2)!k!(n-k)!(2n+2k+3)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{4^{n+2}n!\tan^{2n+2k+5}x}{(2n+4)!k!(n-k)!(2n+2k+5)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}2^{2n+1}n!\sec^{2n+2k+1}x}{(2n+1)!k!(n-k)!(2n+2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}2^{2n+1}n!\sec^{2n+2k-1}x}{(2n+1)!k!(n-k)!(2n+2k-1)}+C$

Harry Peter
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