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Let's say I have integral

Integrate[1/(1 + (f^2/B^2)), {f, 0, Infinity}]

The result is

$$\text{ConditionalExpression}\left[\frac{\pi }{2 \sqrt{\frac{1}{B^2}}},\Im\left(B^2\right)\neq 0\lor \Re\left(B^2\right)\geq 0\right] $$ How can I agree to these conditions and have the Mathematica simplify this result to: $$B\frac{\pi}{2}$$ Or is this possible?

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  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Dec 4 '14 at 16:09
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Just specify the Assumptions for the Integrate:

Integrate[1/(1 + (f^2/B^2)), {f, 0, Infinity}, 
 Assumptions -> B > 0]
(* (B π)/2 *)
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  • $\begingroup$ Perfect! Surfed through help for 30mins but didn't find that info. Also your too fast ;) I can't accept for the next 11mins. $\endgroup$ – ELEC Dec 4 '14 at 16:09
  • $\begingroup$ @ELEC I'd recommend to read other answers too, so you can choose the one, that's suit you the most. $\endgroup$ – m0nhawk Dec 4 '14 at 16:10
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Integrate[1/(1 + (f^2/B^2)), {f, 0, Infinity}, GenerateConditions -> False]
π/(2 Sqrt[1/B^2])
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