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It is known DLMF (25.2.8) that for $\Re s>0$ and for integers $N\geq 1$ $$\zeta(s)=\sum_{k=1}^N\frac{1}{k^s}+\frac{N^{1-s}}{s-1}-s\int_{N}^\infty \frac{x-\lfloor x \rfloor}{x^{s+1}} dx,$$

where $\zeta(s)$ is the Riemann zeta function.

Then if I propose $\rho= \left(\frac{1}{2}+\epsilon \right)+it, $ satisfying $\zeta(\rho)=0$ and such that $0<\epsilon<\frac{1}{2}$ (I don't use this last condition on epsilon), if there are no mistakes one has by direct computations with the real part function that $$\int_{N}^\infty \frac{x-\lfloor x \rfloor}{x^{\frac{3}{2}+\epsilon}} \left( (\frac{1}{2}+\epsilon)\cos (t\log x)+t\sin (t\log x) \right) dx$$ equals to $$\sum_{k=1}^N\frac{\cos (t\log k)}{k^{\frac{1}{2}+\epsilon}}+\frac{N^{\frac{1}{2}-\epsilon}}{(\epsilon-\frac{1}{2})^2+t^2} \left((\epsilon-\frac{1}{2})\cos (t\log N)-t\sin (t\log N) \right). $$ In this context and with the purpose to learn some useful fact I've asked to me some questions. One of those was

Question. Compute $$\int \frac{t\sin at}{b^2+t^2}dt,$$ for real numbers $a> 0$ and $b>0$ (if you consider it, you can use the form $a=\log N$, where $N\geq 2$ is an integer, and $b=\epsilon-\frac{1}{2}$ with $0<\epsilon<\frac{1}{2}$, I have no used this from my evaluation with Wolfram Alpha, and I have no special knowledges about the integral functions $Ci(x)$ and $Si(x)$). Thanks in advance.

My attempt was explore the other integral (the concerning with the cosine function), integration by parts yields as an antiderivative $v=\frac{1}{\epsilon-\frac{1}{2}}\arctan\frac{t}{\epsilon-\frac{1}{2}}$; and using Wolfram Alpha that provide us a result concerning our Question involving $Ci(x), Si(x)$ and hyperbolic functions (it isn't neccesary that you wirte this definition, I will read how are defined). The comment in brackects in the Question, was because I am interested to evaluate integrals of the form $\int_{-T}^T$.

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  • $\begingroup$ If you know previous computations, and you've detected some mistake in my claims please add a comment. Thanks. $\endgroup$ – user243301 May 23 '16 at 8:58
  • $\begingroup$ I've accepted @ClaudeLeibovici 's answer, but are welcome subsequents contributions, comments about my computations. Very thanks much alls users, are perfect! $\endgroup$ – user243301 May 23 '16 at 9:44
  • $\begingroup$ Very thanks much for your edit @mike $\endgroup$ – user243301 May 24 '16 at 8:55
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The trick is to use $$t^2+b^2=(t+i b)(t-ib)$$ and use partial fraction decomposition $$\frac t{t^2+b^2}=\frac{1}{2 (t+i b)}+\frac{1}{2 (t-i b)}$$ So $$\int \frac{t\sin at}{t^2+b^2}dt=\frac{1} 2\int\frac{\sin at}{t+ib}dt+\frac{1} 2\int\frac{\sin at}{t-ib}dt$$ So, now, consider $$I=\int \frac{\sin at}{t+c}dt$$ Change variable $t=u-c$ to get $$I=\int\frac{\sin (a (u-c))}{u}du$$ Develop the sine, make some simple change of variables to arrive to things like $$\int\frac{\sin(x)}x dx=\text{Si}(x)\qquad \int\frac{\cos(x)}x dx=\text{Ci}(x)$$ and you finish with $$I=\cos (a c) \text{Si}(a u)-\sin (a c) \text{Ci}(a u)$$ Consider now the two cases $c=\pm ib$ and play with standard identities.

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  • $\begingroup$ Then very thanks much Claude, I will study your answer. $\endgroup$ – user243301 May 23 '16 at 9:38

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