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I want to evaluate the integral $$\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$$ I tried using Feynman’s trick and introduced the parameter $e^{-\sqrt{1+t^2} x}$, then differentiating with respect to $x$ under the integral. However I don’t see a good way to evaluate the resulting integral. Is there a contour method?

This was an AMM problem ($12145$).

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    $\begingroup$ Is there another substitution that might work better? For example, cos(x * sqrt(1+t^2)). Not sure if this will work, but perhaps it will suggest a useful course of action. I hope this helps. $\endgroup$
    – ad2004
    Nov 25, 2019 at 20:13

2 Answers 2

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You can let $t = \sinh(x)$ to get \begin{align} \int \limits_{-\infty}^\infty \frac{\cos(t) \sin(\sqrt{1+t^2})}{\sqrt{1+t^2}} \, \mathrm{d} t &= \int \limits_{-\infty}^\infty \cos(\sinh(x)) \sin(\cosh(x)) \, \mathrm{d} x = \int \limits_{-\infty}^\infty \frac{1}{2} \left[\sin(\mathrm{e}^{x}) + \sin(\mathrm{e}^{-x})\right] \, \mathrm{d} x \\ &= \int \limits_{-\infty}^\infty \sin(\mathrm{e}^{x}) \, \mathrm{d} x \stackrel{\mathrm{e}^{x} = u}{=} \int \limits_0^\infty \frac{\sin(u)}{u} \, \mathrm{d} u = \frac{\pi}{2} \end{align} using the famous Dirichlet integral and $\sin(a) \cos(b) = \frac{1}{2} [\sin(a+b)+\sin(a-b)]$.

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    $\begingroup$ @emonHR Whenever $\sqrt{1+t^2}$ appears in an integral, this substitution might be good idea (even more so if this term is part of the denominator, since it cancels the $\cosh(x)$ from $\mathrm{d} t = \mathrm{d} \sinh(x)$). $\endgroup$ Nov 25, 2019 at 20:54
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Another solution for an interesting antiderivative. $$I=\int\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt$$

First write $$I=\frac12\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt+\frac12\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt$$ For the first integral, use $u=\sqrt{t^2+1}+t$ to get by the end $$\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt=\int \frac {\sin(u)} u \,du=\text{Si}(u)=\text{Si}(\sqrt{t^2+1}+t)$$ For the second integral, use $u=\sqrt{t^2+1}-t$ to get by the end $$\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt=-\int \frac {\sin(u)} u \,du=-\text{Si}(u)=-\text{Si}(\sqrt{t^2+1}-t)$$

So $$I=\frac 12\Big(\text{Si}\left(\sqrt{t^2+1}+t\right)-\text{Si}\left(\sqrt{t^2+1}-t\right)\Big)$$ For the definite integral $$J=\int_{-p}^p\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt=\text{Si}\left(p-\sqrt{p^2+1}\right)+\text{Si}\left(p+\sqrt{p^2+1}\right)$$

Now, for large values of $p$, composing Taylor series $$p-\sqrt{p^2+1}=-\frac{1}{2 p}+\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$ $$\text{Si}\left(p-\sqrt{p^2+1}\right)=-\frac{1}{2 p}+\frac{19}{144 p^3}+O\left(\frac{1}{p^5}\right)\qquad \to 0$$ $$p+\sqrt{p^2+1}=2 p+\frac{1}{2 p}-\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$ Now, using for large values of $a$ $$\text{Si}(a)=\frac{\pi}{2} -\frac{\left(a^2-2\right) \cos (a)}{a^3}-\frac{\sin (a)}{a^2}+\cdots$$ then the result.

Using $p=100$, the above truncated expressions lead to $1.5733$ while the numerical integration gives $1.5634$.

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