To evaluate this integral here. We're gonna use the limit as a variable approaches infinity instead of leaving the upper limit as infinity. Because we can't direct substitute that. So we'll go ahead and rewrite that as the limit from zero to be. Let me go off to infinity then, uh, to evaluate the integral itself. We're going to use the use substitution if we make you X over to and then do you derivative of X over two is just one half dx. Then isolate the X, multiply the to by both sides. We'll get to do you is equal to DX. So that's what we'll go ahead and replace here for this DX too, do you? Then we could even go ahead and push that to to the front. And then we're really integrating Sign of you, Do you? So the integral what do we have to take the derivative of? To get signed, basically, is the ideas that would have to be negative to, but sorry, Negative too. Code side of you. And we'll go ahead and just plug. Are you right back in X over to leave the limit? The approaches infinity out front and this is going to be from zero Whoops. I wanted to do that red zero to be. And so, as we go ahead and evaluate this now we're going to get negative, too. Co sign, be over to and then minus, uh, negative, too. So it'll turn into a plus to cosign of zero. However, we still have this limit as the approaches infinity out front of here is well, so if we go ahead and apply that now, then this is going to change to cosign of infinity. And because that is undefined, it's just gonna oscillator. Then this whole limit is going to diverge because it's not going to collapse onto one specific value.