A question on the convergence of an integral I'm interested in showing that the integral
$$\lim_{t\to \infty}\int_{1}^{t} \frac{\sin(\theta)}{\theta}\:d\theta$$
converges. I am aware that one can show that $$\int_{0}^{\infty} \frac{\sin(\theta)}{\theta}d\theta=\frac{\pi}{2}$$ using contour integration. But I was wondering if one can prove convergence without necessarily making use of this fact?
Thanks in advance.

You can use Feynman's technique;
$$\int_{0}^{\infty}{\sin(x)\over x}dx\space \rightarrow \space I(a)=\int_{0}^{\infty}{\sin(x)\over x}\cdot e^{-ax}dx$$
$$\therefore {d\over da}I(a)=-\int_{0}^{\infty}{\sin(x)}\cdot e^{-ax}dx$$
$$=-Im\bigg\{\int_{0}^{\infty}e^{-(a-i)x} dx\bigg\}=Im\bigg\{{e^{-(a-i)x }\over(a-i)}{\bigg|_{0}^{\infty}} \bigg\}$$
$$=Im\bigg\{ {-1\over(a-i)}\bigg\}=Im\bigg\{ {-(a+i)\over(a^2+1)}\bigg\}={-1\over(a^2+1)}$$
$$\therefore I(a)=\int{-1\over(1+a^2)}da=-\arctan(a)+C$$
So we have;
$$\int_{0}^{\infty}{\sin(x)\over x}\cdot e^{-ax}dx=-\arctan(a)+C$$
Letting $a\rightarrow \infty$ reveals;
$$0=-{\pi\over2}+C \implies C={\pi\over2}$$
Which gives us our final result;
$$I(0)=\int_{0}^{\infty}{\sin(x)\over x}dx={\pi\over2}$$
