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Here is a possible beginning. Note first that

$$\int_{-1}^1 f(t)^a \cos(at) dt= 2I_a:= 2\int_{0}^1 f(t)^a \cos(at) dt.$$

Hence, it suffices to investigate $I_a(t)$. I_a$. Let me first assume that$f'(t) <0$on$(0,1)$. (Note that if$f'(t_0)=0$for some$t_0\in (0,1)$then$f'(t)=0$on$[0,t_0]$.) This means that the map$t\mapsto f$is one-to-one. We regard$t$as a function of$f$. Then the change in variables formula implies. $$I_a= \int_1^{f(0)} f^a \cos(a t)\frac{dt}{df} df$$ I can make this formula friendlier to the 21st century mathematician by changing notations, $$t \longleftrightarrow \phi,\;\;\; f \longleftrightarrow x$$ and we can rewrite the above as $$I_a= \int_1^{x_0} x^a\frac{d\phi}{dx} \cos( a \phi(x) ) dx = \frac{1}{a}\int_1^{x_0} x^a \frac{d}{dx}\Bigl( \sin\bigl(\; a\phi(x)\;\bigr) \Bigr)dx$$ $$=\frac{1}{a}\Bigl( x^a\sin\bigl( a\phi(x)\bigr)\;\Bigr)\Bigr|^{x_0}_1- \int_1^{x_0}x^{a-1}\sin\bigl(\; a\phi(x) \;\bigr)dx.$$ Now observe that$\phi(x_0) =0$,$\phi(1)=1$, so the first term above goes to zero as$a\to\infty$. At this point it may be useful to look in some books on asymptotics of integrals. A good place to start is Bleistein & Handelsman: Asymptotic expansions of Integrals, Dover Also you need to keep in mind that $$\frac{d\phi}{dx}< 0,\;\;\forall x\in (1,x_0)$$ $$\lim_{x\nearrow x_0} \frac{d\phi}{dx}=-\infty.$$ 4 deleted 1 characters in body Here is a possible beginning. Note first that $$\int_{-1}^1 f(t)^a \cos(at) dt= 2I_a:= 2\int_{0}^1 f(t)^a \cos(at) dt.$$ Hence, it suffices to investigate$I_a(t)$. Let me first assume that$f'(t) <0$on$(0,1)$. (Note that if$f'(t_0)=0$for some$t_0\in (0,1)$then$f'(t)=0$on$[0,t_0]$.) This means that the map$t\mapsto f$is one-to-one. We regard$t$as a function of$f$. Then the change in variables formula implies. $$I_a= \int_1^{f(0)} f^a \cos(a t)\frac{dt}{df} df$$ I can make this formula friendlier to the 21st century mathematician by changing notations, $$t \longleftrightarrow \phi,\;\;\; f \longleftrightarrow x$$ and we can rewrite the above as $$I_a= \int_1^{x_0} x^a\frac{d\phi}{dx} \cos( a \phi(x) ) dx = \frac{1}{a}\int_1^{x_0} x^a \frac{d}{dx}\Bigl( \sin\bigl(\; a\phi(x)\;\bigr) \Bigr)dx$$ $$=\frac{1}{a}\Bigl( x^a\sin\bigl( a\phi(x)\bigr)\;\Bigr)\Bigr|^{x_0}_1- \int_1^{x_0}x^{a-1}\sin\bigl(\; a\phi(x) \;\bigr)dx.$$ Now observe that$\phi(x_0) =0$,$\phi(1)=1$, so the first term above goes to zero as$a\to\infty$. At this point it may be useful to look in some books on asymptotics of integrals. A good place to start is Bleistein & Handelsman: Asymmptotic Asymptotic expansions of Integrals, Dover Also you need to keep in mind that $$\frac{d\phi}{dx}< 0,\;\;\forall x\in (1,x_0)$$ $$\lim_{x\nearrow x_0} \frac{d\phi}{dx}=-\infty.$$ 3 added 1 characters in body Here is a possible beginning. Note first that $$\int_{-1}^1 f(t)^a \cos(at) dt= 2I_a:= \int_{0}^1 2\int_{0}^1 f(t)^a \cos(at) dt.$$ Hence, it suffices to investigate$I_a(t)$. Let me first assume that$f'(t) <0$on$(0,1)$. (Note that if$f'(t_0)=0$for some$t_0\in (0,1)$then$f'(t)=0$on$[0,t_0]$.) This means that the map$t\mapsto f$is one-to-one. We regard$t$as a function of$f$. Then the change in variables formula implies. $$I_a= \int_1^{f(0)} f^a \cos(a t)\frac{dt}{df} df$$ I can make this formula friendlier to the 21st century mathematician by changing notations, $$t \longleftrightarrow \phi,\;\;\; f \longleftrightarrow x$$ and we can rewrite the above as $$I_a= \int_1^{x_0} x^a\frac{d\phi}{dx} \cos( a \phi(x) ) dx = \frac{1}{a}\int_1^{x_0} x^a \frac{d}{dx}\Bigl( \sin\bigl(\; a\phi(x)\;\bigr) \Bigr)dx$$ $$=\frac{1}{a}\Bigl( x^a\sin\bigl( a\phi(x)\bigr)\;\Bigr)\Bigr|^{x_0}_1- \int_1^{x_0}x^{a-1}\sin\bigl(\; a\phi(x) \;\bigr)dx.$$ Now observe that$\phi(x_0) =0$,$\phi(1)=1$, so the first term above goes to zero as$a\to\infty\$.

At this point it may be useful to look in some books on asymptotics of integrals. A good place to start is

Bleistein & Handelsman: Asymmptotic expansions of Integrals, Dover

Also you need to keep in mind that

$$\frac{d\phi}{dx}< 0,\;\;\forall x\in (1,x_0)$$

$$\lim_{x\nearrow x_0} \frac{d\phi}{dx}=-\infty.$$

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