Residue theorem problem involving real value Intergral

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The integral from infinity to 0 of x^2/(x^2+1)(x^2+16) is evaluated using the residue theorem, focusing on the singularities at i and 4i, which are simple poles. The residues at these poles are calculated, yielding -2i/15 for 4i and i/30 for i, resulting in a total residue of -3i/30. Since the integral is taken from 0 to negative infinity, the final result is multiplied by πi, leading to a value of π/10. The discussion emphasizes the importance of understanding each step in the calculation process. The solution appears to be correct based on the residue theorem application.
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Homework Statement



∫ from infinity to 0 of x^2/(x^2+1)(x^2+16)

Homework Equations



The Integral will be the sum of the residues times 2πi

The Attempt at a Solution



The only singularities that matter are at i and 4i and they are simple poles. So x^2/(x^2+1)/(2x) and I plug in 4i and get -2i/(15) then I plug in i into x^2/(x^2+16)/(2x) and get i/30 and when I add them I get -3i/30. Because it's only from 0 and to negative infinity I mult that by ∏i and I get ∏/10.
 
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Looks good.

You glossed over several steps, which I'm assuming you understand. Just make sure you know why you can do what you did.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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