Htykuut

which one of the following hold For all continuous function $f : [-pi ,pi] rightarrow mathbb{C}$

$a)$ If $f(-t) =-f(t)$ for each $t in [-pi,pi]$,then $int_{-pi}^{pi}f(t)dt=0$

$b)$ $int_{-pi}^{pi}f(-t)dt=- int_{-pi}^{pi}f(t)dt $

My attempt : I thinks option b) will hold

and option a) will not hold because $int_{-pi}^{pi}f(t)dt=2int_{0}^{pi}f(t)dt neq 0$

Is it correct ?

Actually both option you have given are equivalent. (Check b if you have some typos)

And both are correct . As function is odd so if you integrate it you get even function and by subtracting you get 0

option b) is false

$displaystyle underset{[-1,1]}inf fint_{-1}^{1}dtleint_{-1}^{1}f(t)dtle underset{[-1,1]}sup fint_{-1}^{1}dtiff 2underset{[-1,1]}inf fleint_{-1}^{1}f(t)dtle 2underset{[-1,1]}sup f$

so $f: xto cos^2(x)$ is counterexample

$displaystyle int_{-1}^{1}cos^2(t)dt =int_{0}^{1}1+cos(2t)dt=bigg[t+dfrac{sin(2t)}{2}bigg]^1_0=1+dfrac{sin2 }{2}>1ge f(x)ge0,quad forall xinmathbb{R}$

It looks on the face of it that the other answer (the first you've got - this is the second) is just plainly obviously correct ... but in the abstract theory of functions I have seen so many ingeniously constructed 'pathological' functions that somehow manage to defy & foil what appear on the face of it to be the plainest of truths that I wouldn't even venture say with absolute certainty that $x=-ximplies x=0$.

Update

You have to be really careful looking at these, don't you, as it's so easy to make a little slip.

As for (a), it's just saying that the function is odd ... and therefore that the integral is 0, because every $operatorname{f}(-t)dt$ is cancelled by a $operatorname{f}(+t)dt$

But then (b) is stated independently of (a): if you're not careful, you can presume that the condition in (a) carries over into (b). And then another point you have to be careful about is that $$int_{-pi}^pioperatorname{f}(-t)dt$$ is not $$int_{-pi}^pioperatorname{f}(t)dt$$ with $-t$ substituted for $t$, but simply the integral of the function $operatorname{g}(t)$ obtained by reflecting $operatorname{f}(t)$ about $t=0$ ... meaning that in doing the integral it is the same integral but just carried out in reverse order. So (b) does not hold in general, but only when (a) applies ... in which case it's true by reason only of the being zero of the integral.

All this of course precludes any consideration of the sort of pathological function I was talking about at first, that can spring allmanner of surprise. But that is probably precluded by the continuity requirement anyway.