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### Solution

The convergence of the improper integral $$\int_{0}^{\infty} \frac{\sin (t)}{t} d t$$ is classic : There is no problem at the origin because $$t \mapsto \sin (t) / t$$ is extendable by continuity, the only problem is therefore in $$\infty$$. integrable on $$[0,1]$$, it is sufficient to ensure convergence on $$[1,+\infty[:$$ let $$x>1$$, an integration by parts gives $$\int_{1}^{x} \frac{\sin (t)}{t} d t=\left[-\frac{\cos (t)}{t}\right]_{1}^{x}-\int_{0}^{x} \frac{\cos (t)}{t^{2}} d t$$ when $$x$$ approaches $$+\infty$$ the term $$«$$ in brackets $$»$$ approaches $$\sin (1)$$ and the second (since $$\left|\cos (t) / t^{2}\right| \leq t^{-2} \in L^{1}\left([1,+\infty[))\right.$$ toward $$\int_{1}^{\infty} \cos (t) / t^{2} d t \in \mathbb{R}$$. Therefore $$\int_{0}^{+\infty} \frac{\sin (t)}{t} d t$$ converges.
Let
$I(b)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-b x} d x\;\;\;\;\;\;\;(1)$
$I(0)=\int_{0}^{\infty} \frac{\sin x}{x} \cdot e^{-0 x} d x$
$=\int_{0}^{\infty} \frac{\sin (x)}{x} d x\;\;\;\;\;\;\;(2)$
Now,
$I(b)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-b x} d x$
Differentiate resp $$b$$ :
\begin{aligned} I^{\prime}(b) &=\int_{0}^{\infty} \frac{d}{d b}\left(\frac{\sin (x)}{x} e^{-b x}\right) d x \\\\ I^{\prime}(b) &=\int_{0}^{\infty}-\frac{x \sin (x)}{x} e^{-b x} d x \\\\ I^{\prime}(b) &=-\int_{0}^{\infty} \sin (x) e^{-b x} d x . \end{aligned}
Integrating by parts
$I^{\prime}(b)=-\left[\frac{-e^{-b x}(\cos x+b \sin x)}{b^{2}+1}\right]_{x=0}^{x=\infty}$
Applying limits
\begin{aligned} I^{\prime}(b) &=0- \frac{e^{-b 0}(\cos 0+b \sin 0)}{b^{2}+1}\\\\ I^{\prime}(b) &=\frac{-1}{b^{2}+1} \end{aligned}
Integrating resp $$b$$, we get :
\begin{aligned} &\int I^{\prime}(b) d b=\int-\frac{1}{b^{2}+1} d b \\\\ &I(b)=-\tan ^{-1}(b)+c=\int_{0}^{\infty} \frac{\sin x\; e^{-b x}}{x} d x\;\;\;\;\;\;from\;(1) \end{aligned}
\begin{aligned} &\text { Let } b \rightarrow \infty \\\\ &-\tan ^{-1}(\infty)+c=\int_{0}^{\infty} \frac{\sin x}{x} e^{-\infty .x} d x \\\\ &\frac{-\pi}{2}+c=0 \Rightarrow c=\pi / 2 \\\\ &\therefore I(b)=-\tan ^{-1}(b)+\pi / 2 \\\\ &I(0)=-\tan ^{-1}(0)+\pi / 2 \Rightarrow I(0)=\pi / 2 \\\\ &\text { From (2), } I(0)=\int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2} \end{aligned}
Home -> Solved problems -> Calculate the integral

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