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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Sign up to join this communityI am a beginner with Mathematica, and I would like to know if it is possible to calculate this kind of integral:
$$ \int\limits_0^{0.8}\cos^2 \left(\frac{\pi z}{1.6} \right)\int\limits_z^{\infty}\frac{e^{-3.367u}}{u^2}du\ dz $$
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}]]
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]
d something
at the end. Is this mathematically even valid? $\endgroup$ – Nasser Dec 21 '17 at 14:45Integrate
orNIntegrate
yet? $\endgroup$ – march Dec 21 '17 at 21:03