Home -> Solved problems -> Calculate the integral

### Solution

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 \(b\) :

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

Integrating resp \(b\), we get :

Home -> Solved problems -> Calculate the integral

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Why does the number \(98\) disappear when writing the decimal expansion of \(\frac{1}{9801}\) ?

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