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Trigonometric integral

Prove that

    \[\bigintsss_0^{\infty}\frac{\sin \sqrt{x^2+1} \cos x}{\sqrt{x^2+1}} \, \mathrm{d}x=\frac{\pi}{4}\]

Solution

Let \mathcal{J} be the integral. Note that

    \[2 \sin \sqrt{x^2+1} \cos x =  \sin \left ( \sqrt{1+x^2} - x \right ) + \sin \left ( \sqrt{1+x^2}+x \right ) \]

and hence:

    \[2\mathcal{J} = \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x + \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} + x \right )}{\sqrt{1+x^2}} \, \mathrm{d}x\]

For the integral \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x we apply the substitution t \mapsto \sqrt{1+x^2}-x. Then, x = \frac{1-t^2}{2t} and

(1)   \begin{equation*} \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x = \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

and similarly by applying the change of variables t \mapsto \sqrt{1+x^2} + x at the second integral we get that

(2)   \begin{equation*} \int_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} \right )}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

Adding equations (1) , (2) we get that

    \begin{align*} 2\mathcal{J} &= \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t + \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \int_{0}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \frac{\pi}{2} \end{align*}

and the result follows.

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