# uniform bound for sine integral function

Prove that for any $0<a<b$, $$\left|\int_a^b\frac{\sin x}{x}\,dx\right|\le4$$ Here is my approach. I used integration by parts to prove that LHS is bounded by $3$ when $a\ge 1$. I will be done if I can show LHS is $\le1$ when $b=1$. I do not know if this is true or not.

• [You can find usefull this answering][1] Remember that $\frac{sinx}{x}$ is an odd function. [1]: math.stackexchange.com/questions/5248/… – Skills Feb 10 '15 at 23:44
• @DavideMarano Here $a>0$. So the fact the integrand is odd is not helpful. – Tony B Feb 11 '15 at 23:53

Hint: Prove that the integral achieves its maximum value when $a = -\pi$, $b = \pi$. Then just complete the proof for that case.
• Did you mean $a=0, b=\pi$? The function is odd. – Tony B Feb 11 '15 at 23:44
This is a bit old, but anyway. As for the question for the LHS being $\leq 1$ when $b =1$, I think one can prove directly that $$\int_0^1\frac{\sin x}{x}dx \leq 1. \quad(*)$$ Integrating by parts $$\int_\epsilon^1\frac{\sin x}{x}dx = -\sin\epsilon\ln\epsilon - \int^1_\epsilon\ln x\cos x dx.$$ Then use that $$- \int^1_\epsilon\ln x\cos x dx \leq - \int^1_\epsilon\ln xdx,$$ because $-\ln x\cos x\leq -\ln x$ for $\epsilon < x < 1$ and evaluate the integral to obtain $$\int_\epsilon^1\frac{\sin x}{x}dx \leq -\sin\epsilon\ln\epsilon + 1 - \epsilon + \epsilon\ln\epsilon.$$ Finally, take the limit $\epsilon \to 0$ (L'Hôpital, for example) to get $(*)$.