Does anyone know which textbook this question is from? $\displaystyle \int_0^x \frac{\sin t}{t+1}dt > 0$ for all $x>0$

Question

So I'm currently doing integration problems preparing for final exam, and I would like to practice problems like the following:

Prove that $\displaystyle \int_0^x \frac{\sin t}{t+1}dt > 0 \ \ $ $\forall x>0$

Can anyone tell me which textbook it is from?

Answer 1

The integral involves special functions known as trigonometric integral, but the problem become quite trivial after you know those functions. Those functions are introduced in most books under the name "mathematical methods for physicists" or similar books for other fields.

Are you taking a introductory calculus course or more advanced?

Answer 2

There is no universal textbook for problem like these. Below is how you would prove such a problem.

Let $m$ be the largest integer such that $2\pi m \leq x$. We then have $$\int_0^x = \int_{2m\pi}^x + \sum_{n=0}^{(m-1)} \int_{2 n \pi}^{(2n+2)\pi}$$ It suffice to show that $$\int_{2m\pi}^x \dfrac{\sin(t)}{t+1} dt > 0 \text{ and } \int_{2 n \pi}^{(2n+2)\pi} \dfrac{\sin(t)}{t+1} dt > 0$$ Further, note that $$\int_{2 n \pi}^{(2n+2)\pi} \dfrac{\sin(t)}{t+1} dt > 0 \implies \int_{2m\pi}^x \dfrac{\sin(t)}{t+1} dt > 0 \tag{$\star$}$$ $(\star)$ is because if $x \in [2m\pi, (2m+1)\pi)$, the integrand is positive and if $x \in [(2m+1)\pi,(2m+2)\pi)$, then $\displaystyle \int_{2m\pi}^x \dfrac{\sin(t)}{t+1} dt > \int_{2 m \pi}^{(2m+2)\pi} \dfrac{\sin(t)}{t+1} dt$.

Hence, it suffice to prove that $$\int_{2 n \pi}^{(2n+2)\pi} \dfrac{\sin(t)}{t+1} dt > 0$$ \begin{align} \int_{2 n \pi}^{(2n+2)\pi} \dfrac{\sin(t)}{t+1} dt & = \int_{2n \pi}^{(2n+1)\pi}\dfrac{\sin(t)}{t+1} dt + \int_{(2n+1) \pi}^{(2n+2)\pi}\dfrac{\sin(t)}{t+1} dt\\ & = \int_{2n\pi}^{(2n+1)\pi} \underbrace{\left(\dfrac1{t+1} - \dfrac1{t+\pi+1} \right)\sin(t)}_{>0} dt \end{align} where the last step is obtained by changing the variable $t \to x+\pi$ in the second integral and hence proved.

Answer 3

This is from Apóstol's Calculus Vol I. I don't have it with me right now, I can update with the precise page.

ADD A slick solution is to integrate by parts, letting $u=1-\cos t$, $v=(1+t)^{-1}dt$ so that

$$\begin{align}\int_0^x vdu=\int_0^x\frac{\sin t}{1+t}dt &=uv\mid_0^x-\int_0^x udv\\&=\frac{1-\cos x}{1+x}+\int_0^x\frac{1-\cos t}{(1+t)^2}dt>0\end{align}$$