How to show $\lim_{t \rightarrow 0^+} \int_0^\infty \dfrac{\sin x}{x + t}dx = \dfrac{\pi}{2}$?

Question

It is well-known that $$ \int_0^\infty \dfrac{\sin x}{x}dx = \dfrac{\pi}{2} $$ Let $$ f(t) = \int_0^\infty \dfrac{\sin x}{x + t}dx \ \forall t > 0 $$ I want to know if $$ \lim_{t \downarrow 0} f(t) = \dfrac{\pi}{2} $$

From Wolfram Alpha, this is indeed true. However, the answer it gives is via the cosine integral function and sine integral function. I wonder if whether we can show this solely by convergence theorem. I tried to bound the integrand by $\left\vert \dfrac{\sin x}{x} \right\vert$ but this is not Riemann integrable on $(0, \infty)$. Any hints would be appreciated, thank you

Answer 1

Here's the full answer based on @RobertZ's comment. We have $$ \begin{align*} \left\vert f(t) - \int_0^\infty \dfrac{\sin x}{x}dx \right\vert &= \left\vert t\int_0^\infty \dfrac{\sin x}{x(x + t)}dx \right\vert\\ &\le t \int_0^1 \left\vert \dfrac{\sin x}{x} \right\vert \dfrac{1}{x + t}dx + t \int_1^\infty \vert \sin x \vert \dfrac{1}{x(x + t)}dx\\ &\le t \int_0^1 \dfrac{dx}{x + t} + t \int_1^\infty \dfrac{dx}{x(x+t)}\\ &= t \ln\left(\dfrac{t + 1}{t}\right) + \ln(1 + t) \xrightarrow{t \downarrow 0} 0 \end{align*} $$ Therefore, $$ \lim_{t \downarrow 0} \int_0^\infty \dfrac{\sin x}{x + t}dx = \int_0^\infty \dfrac{\sin x}{x}dx = \dfrac{\pi}{2} \hspace{0.5cm}\blacksquare $$

Answer 2

Observe that

$$f(t)=\int_{t}^\infty\frac{\sin(x-t)}{x}~\mathrm{d}x=\cos t\int_{t}^\infty\frac{\sin x}{x}~\mathrm{d}x-\sin t\int_t^\infty\frac{\cos x}{x}~\mathrm{d}x.$$

(this is justified as the second integral converges for $t>0$). Now

$$\lim_{t\downarrow0}\int_t^\infty\frac{\sin x}{x}~\mathrm{d}x=\int_0^\infty\frac{\sin x}{x}~\mathrm{d}x$$

(this can easily be justified by splitting the integral at $1$ and applying the DCT for example), and furthermore

$$\left\lvert\sin t\int_t^1\frac{\cos x}{x}~\mathrm{d}x\right\rvert\leq\lvert\sin t\rvert\int_t^1\frac{\mathrm{d}x}{x}=\left\lvert\sin t\ln t\right\rvert\to0$$

as $t\downarrow 0$, so that

$$\sin t\int_t^\infty\frac{\cos x}{x}~\mathrm{d}x=\sin t\int_t^1\frac{\cos x}{x}~\mathrm{d}x+\sin t\int_1^\infty\frac{\cos x}{x}~\mathrm{d}x\to0$$

as $t\downarrow0$. It follows that

$$\lim_{t\downarrow0}f(t)=\int_0^\infty\frac{\sin x}{x}~\mathrm{d}x=\frac{\pi}{2}.$$

Answer 3

If you want to go beyond the limit $$f(t) = \int_0^\infty \dfrac{\sin (x)}{x + t}\,dx=\text{Ci}(t) \sin (t)+\frac{1}{2} (\pi -2 \text{Si}(t)) \cos (t) $$ where appear the sine and cosine integral functions.

Expanded as a series around $t=0$ $$f(t)=\frac{\pi }{2}+ (\log (t)+\gamma -1)\,t+O\left(t^2\right)$$