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I am a beginner with Mathematica, and I would like to know if it is possible to calculate this kind of integral:

$$ \int\limits_0^{0.8}\cos^2 \left(\frac{\pi z}{1.6} \right)\int\limits_z^{\infty}\frac{e^{-3.367u}}{u^2}du\ dz $$

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  • $\begingroup$ I see two integrals signs, but only one d something at the end. Is this mathematically even valid? $\endgroup$ – Nasser Dec 21 '17 at 14:45
  • $\begingroup$ already corrected, thanks! $\endgroup$ – Rafael Dec 21 '17 at 14:48
  • $\begingroup$ Did you check the documentation for Integrate or NIntegrate yet? $\endgroup$ – march Dec 21 '17 at 21:03
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You can evaluate it like this, but it blows up at z=0

First evaluate the inner integral on its own

 int1=Assuming[z>= 0,Integrate[  Exp[-3367/1000 u]/u^2,{u,z,Infinity}]]

Mathematica graphics

You see that it has to be for z>0 now evaluate the complete integral

 Integrate[ Cos[ Pi z/(16/10)]^2* int1 ,{z,0,8/10},PrincipalValue->True]

Mathematica graphics

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