# Math Help - arctan(-infinite)

1. ## arctan(-infinite)

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

is that really just -infinite , or 0??

2. Originally Posted by Legendsn3verdie
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

3. Originally Posted by Mathstud28
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??

4. Originally Posted by Legendsn3verdie
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

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