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Let $0<a<b$ and $f: \mathbb{R} \to \mathbb{R}$, $f(x) = \int_a^b \frac{sin(xt)}{t} \,dt$, $\forall x \in \mathbb{R}$.

Prove that $f$ is strictly increasing on $[0,\frac{\pi}{a+b}]$.

Attempt:

Looking at $k < l$, $k,l, \in [0,\frac{\pi}{a+b}]$, I've found some restrictions on $a,b$ on, mainly $a>1$, implies that $\int_a^b f(l) \,dt - \int_a^b f(k) \,dt < 0$. But, in general, that integral is not less than $0$. So, now we only have to consider $b \le 1$. And clearly this is where the fact that $x \in [0,\frac{\pi}{a+b}]$ comes into play. But I can't quite grasp how this restriction helps.

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  • $\begingroup$ what would differentiating wrt $x$ give? $\endgroup$ – Maesumi Dec 5 '13 at 22:35
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Hint:

Put $u=xt$ then $f(x)=\int_{ax}^{bx} \frac{\sin u}{u} du$, and, for all $x >0$: \begin{eqnarray*}f'(x)&=&b \frac{\sin(bx)}{bx}-a \frac{\sin (ax)}{ax} \\ &=&\frac 1x (\sin (bx)-\sin(ax)) \\ &=&\frac{2}{x} \cos\frac{a+b}{2}x \sin \frac{b-a}{2}x \end{eqnarray*}

Now if $0 \leq x \leq \frac{\pi}{a+b}$ then , we have : $$0 \leq \frac{a+b}2 x \leq \frac{\pi}2$$ and $$0 \leq \frac{b-a}2 x \leq \frac{b-a}{b+a}\frac{\pi}2 \leq \frac{\pi}{2}$$ and sow : $f'(x) \geq 0$.

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  • $\begingroup$ I'm not quite sure how you derived $f'(x)$? It looks like the fundamental theorem of calculus, but applying FTC with the u-substituted form yields $\frac{sin(xb)}{xb} - \frac{sin(xa)}{xa}$. Of course, this is $f'(u)$. But, converting back to the variable x via $f'(\frac{u}{t})$ doesn't quite make sense to me? (Also, in the second line of deriving $f'(x)$, I believe you mean $sin(bx) - sin(ax)$, not $sin(ax) - sin(bx)$) $\endgroup$ – user110669 Dec 6 '13 at 0:42
  • $\begingroup$ if $F(x)=\int_c^x f(t) dt$ and $f$ continuou then $F'(x)=f(x)$. Now $G(x)=\int_c^{ax} f(t) dt$ gives $G(x)=F(ax)$; by composition $G'(x)=a F'(ax)=af(ax)$ $\endgroup$ – Mohamed Dec 6 '13 at 1:27

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