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We have the integral :

$$I(t)=-i\int_0^\infty \frac{\log\left[\frac{\sin(t\log\sqrt{1+ix})}{\log(1+ix)} \right ]-\log\left[\frac{\sin(t\log\sqrt{1-ix})}{\log(1-ix)} \right ]}{e^{2\pi x}-1} \, dx$$

I have tried everything to compute the integral, but it seems it's not doable in terms of elementary functions. For instance, the form of integral suggests that the Able-Plana formula can be used, but it can't. And closing the contour is troublesome. I have reasons to believe that the integral has logarithmic singularities, and can be expressed as: $$I(t)=f(t)+\sum_{\beta_{j}}\log\left(1-\frac{t^{2}}{\beta_{j}^{2}}\right)$$ Where $f(t)$ is an even,entire function -possibly zero- and the numbers $\beta_{j}$ are positive, real numbers. However, i haven't been able to prove that. A plot of the function (numerical integration) could be helpful.

EDIT

We can express the integral as : $$\int_{1-i\infty}^{1+i\infty}\frac{\log \left[\frac{ \sin{\left(\frac{t}{2}\log{u}\right)}}{\log{u}} \right ]}{e^{2\pi i u}-1}du-\int_{1}^{1+i\infty}\log \left[\frac{ \sin{\left(\frac{t}{2}\log{u}\right)}}{\log{u}} \right ]du$$

EDIT 2

The derivative of the integral can be expressed as : $$\frac{d}{dt}I(t)=\frac{t}{2i}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\frac{\log(1+ix)^{2}}{\left(\frac{t}{2}\log(1+ix) \right )^{2}-\pi ^{2}n^{2}} -\frac{\log(1-ix)^{2}}{\left(\frac{t}{2}\log(1-ix) \right )^{2}-\pi ^{2}n^{2}} }{e^{2\pi x}-1}$$ We can do the integral if we can calculate $$\int_{0}^{\infty}\frac{\log(1\pm ix)^{2}}{\left(\frac{t}{2}\log(1 \pm ix) \right )^{2}-\pi ^{2}n^{2}}e^{-2\pi mx}dx$$

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  • $\begingroup$ what a base of log? try to simplify integral,hint,log(y)-log(x)=log(x/y) $\endgroup$ – haqnatural Jul 8 '15 at 0:25
  • $\begingroup$ the base is $e$, your hint doesn't make sense ... the integral $$-i\int_{0}^{\infty}\frac{\log\left[\log(1+ix) \right ]-\log\left[\log(1-ix) \right ]}{e^{2\pi x}-1}dx$$ doesn't converge ! $\endgroup$ – Mohammad Al Jamal Jul 8 '15 at 0:38
  • $\begingroup$ The question might be easier if we had $\sin( t\sqrt{\ln(1+ix)})$ instead of $\sin(\frac{t}{2}\ln(1+ix))$ $\endgroup$ – will Jul 13 '15 at 21:01
  • $\begingroup$ can you explain in more details ? $\endgroup$ – Mohammad Al Jamal Jul 14 '15 at 18:14
  • $\begingroup$ Euler used an infinite product: $sin(\pi z) = \pi z\prod(1-z^2n^{-2})$, which looks similar to the lograthmic singularities you expected. Upon further reflection, the square root is almost irrelevant. $\endgroup$ – will Jul 14 '15 at 19:02
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As discussion

This is seem to me is of Bromwich type integral.

\begin{align} I\left( t \right) &= \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right) - \log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 - ix} } \right)} \over {\log \left( {1 - ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} \\ &= \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} - \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 - ix} } \right)} \over {\log \left( {1 - ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} \\ &= \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} - \int_0^{ - \infty } {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{e^{ - 2\pi y} - 1}}d\left( { - y} \right)} \\ &= \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} - \int_{ - \infty }^0 {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{e^{ - 2\pi y} - 1}}dy} \\ &= \int_0^\infty {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)}}{{e^{2\pi x} - 1}}dx} + \int_{ - \infty }^0 {\frac{{\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)}}{{1 - e^{ - 2\pi y} }}dy} \end{align} where we used the substitution $-x=y$ in the second integral.

Now, using geometric series we have \begin{align} \frac{1}{{e^{2\pi x} - 1}} = \frac{{e^{ - 2\pi x} }}{{1 - e^{ - 2\pi x} }} = e^{ - 2\pi x} \sum\limits_{n = 0}^\infty {e^{ - 2n\pi x} } = \sum\limits_{n = 0}^\infty {e^{ - 2\left( {n + 1} \right)\pi x} } \end{align} so that \begin{align} \int_0^\infty {\left( {\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + ix} } \right)} \over {\log \left( {1 + ix} \right)}}}} \right)} \right)\sum\limits_{n = 0}^\infty {e^{ - 2\left( {n + 1} \right)\pi x} } dx} + \int_{ - \infty }^0 {\log \left( {{\textstyle{{\sin \left( {t\log \sqrt {1 + iy} } \right)} \over {\log \left( {1 + iy} \right)}}}} \right)\sum\limits_{n = 0}^\infty {e^{ - 2n\pi y} } dy} \end{align} Let $ u = \sqrt {1 + iy} \Rightarrow u^2 = 1 + iy \Rightarrow 2udu = idy \Rightarrow dy = - 2iudu $ \begin{align} I(t):= - 2i\int_1^{1 + i\infty } {\left( {\log \left( {{\textstyle{{\sin \left( {t\log u } \right)} \over {\log \left( {u^2} \right)}}}} \right)} \right)\sum\limits_{n = 0}^\infty {e^{ 2i\left( {n + 1} \right)\pi \left( {u^2 - 1} \right) } } udu} \\ &- 2i\int_{1 - i\infty }^1 {\log \left( {{\textstyle{{\sin \left( {t\log u } \right)} \over {\log \left( {u^2} \right)}}}} \right)\sum\limits_{n = 0}^\infty {e^{ 2in\pi \left( {u^2 - 1} \right) } } udu} \end{align} Now use the fact that: Theorem: Let $F$ be an analytic dunction whose singularities $z_1,z_2,\cdots,z_n$ belong to the half-plane $\{{z|\Re({z})<c}\}$ and let $ \mathop {\lim }\limits_{z \to \infty } F\left( z \right) = 0$. Then \begin{align} \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{zt} F\left( z \right)dz} = \sum\limits_{k = 1}^n {\mathop {{\mathop{\rm Res}\nolimits} }\limits_{z = z_k } \left\{ {e^{zt} F\left( z \right)} \right\}} \end{align}

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  • $\begingroup$ there are problems in your reasoning ... $\frac{1}{e^{-2\pi y}-1}$ should be expanded when $y<0$. $\endgroup$ – Mohammad Al Jamal Jul 8 '15 at 22:03
  • $\begingroup$ Since $x \in [0, \infty)$ we have assumed that $-x=y$ which is $<0$? Tell if am I wrong? $\endgroup$ – Mohammad W. Alomari Jul 8 '15 at 23:37

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