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Math Help - arctan(-infinite)

  1. #1
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    arctan(-infinite)

    what is it?? we are doing improper integrals and i just got a arctan(-infinite)

    is that really just -infinite , or 0??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    what is it?? we are doing improper integrals and i just got a arctan(-infinite)

    is that really just -infinite , or 0??
    Two things, the arctangent function is odd and has the relation

    \arctan\left(x\right)=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)

    I think you can see what do to
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Two things, the arctangent function is odd and has the relation

    \arctan\left(x\right)=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)

    I think you can see what do to
    so ur saying its zero, i just figured its -infintie because i remeber that the bounds of arctan from pie/2 to -pie/2 are undifined and for arcsin its boounds are restricted at pie/2 to -pie/2.. so if u take ur ti 83 an put in a really big negative number for arctan it keeps going.. but if u do the same for arc sin it doesnt exsist.. so i conclue arctan-infite = -infinite.


    is this righT??
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    so ur saying its zero, i just figured its -infintie because i remeber that the bounds of arctan from pie/2 to -pie/2 are undifined and for arcsin its boounds are restricted at pie/2 to -pie/2.. so if u take ur ti 83 an put in a really big negative number for arctan it keeps going.. but if u do the same for arc sin it doesnt exsist.. so i conclue arctan-infite = -infinite.


    is this righT??

    No

    \frac{-\pi}{2}\leqslant\arctan(x)\leqslant\frac{\pi}{2}
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