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In the Gradshteyn and Ryzhik Table of Integrals, the following integral appears (3.876.1, page 486 in the 8th edition): \begin{equation} \int_0^{\infty} \frac{\sin (p \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}} \cos (bx) dx = \begin{cases} \frac{\pi}{2} J_0 \left( a \sqrt{p^2 - b^2} \right) & 0 < b < p \\ 0 & b > p > 0 \end{cases} \end{equation} for $a > 0$, where $J_0$ is the Bessel function.

I am interested how this result could be derived (not necessarily rigorously proved), especially the range of $p$ such that the value is $0$.

Thank you.

Maximilian Janisch
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questions
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  • No it's not, check the $\sin$. Although it is not hard to do a change of variable for $p > 0$ and effectively normalize it to 1, so it's not too relevant. – questions Jul 28 '15 at 02:28
  • in my edition, the note on bibliogrphay notation is the final page with roman numerals...ET I 26(30) is volume I of Erdelyi et al, Tables of Integral Transforms (1954), page 26, formula numbered (30). In turn, that may refer to something earlier. http://authors.library.caltech.edu/43489/1/Volume%201.pdf – Will Jagy Jul 28 '15 at 03:19
  • Let $x=a\sinh t$, followed by the trigonometric formula for $\sin A\cos B$, further simplifications, and, ultimately, the integral expressions for Bessel functions. – Lucian Jul 28 '15 at 04:47

1 Answers1

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$$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\ $$

Enforcing the substitution $x\to a\sinh x$ and assuming that $a>0$ yields

$$\begin{align} I(a,b,p)&=\int_0^{\infty}\sin\left(pa\cosh x\right)\cos(ba\sinh x)\,dx\\\\ &=\frac12\int_0^{\infty}\left(\sin\left(pa\cosh x+ba\sinh x\right)+\sin\left(pa\cosh x-ba\sinh x\right)\right)\,dx\\\\ &=\frac12\int_{-\infty}^{\infty}\sin\left(pa\cosh x+ba\sinh x\right)\,dx \end{align}$$

Recalling that

$$A\cosh x+B\sinh x= \begin{cases} \sqrt{A^2-B^2}\cosh(x-\text{artanh}(B/A))&,0<B<A\\\\ \sqrt{B^2-A^2}\sinh(x-\text{artanh}(A/B))&,B>A>0 \end{cases} $$

we have

$$I(a,b,p)= \begin{cases} \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\ \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{b^2-p^2}\sinh x\right)\,dx&,b>p>0 \end{cases} $$

Noting that the integrand of the first integral is an even function of $x$, while the integrand of the second integral is an odd function of $x$ reveals

$$I(a,b,p)= \begin{cases} \int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx&,0<b<p\\\\ 0&,b>p>0 \end{cases} $$

From Equation $(10.9.9)$ HERE, we see that

$$\int_{0}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right)\,dx=\frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)$$

for $a\sqrt{p^2-b^2}>0$ and $p>b>0$

whereupon we have the final result

$$I(a,b,p)= \begin{cases} \frac{\pi}{2}J_0\left(a\sqrt{p^2-b^2}\right)\,dx&,0<b<p\\\\ 0&,b>p>0 \end{cases} $$

for $a>0$.

Mark Viola
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  • Very nice ! (+1) To make the answer even more complete, you could add a proof of 10.9.9 if you can...?! – tired Jul 28 '15 at 13:47
  • @Dr.MV This exactly answers my question, thank you! If possible/you are aware (I know this was not part of the original question) could you explain "why" there is such singular behavior at $p = b$? Is there some intuition that could have hinted such behavior? If not, I will gladly accept the answer as is. – questions Jul 28 '15 at 15:35
  • @questions You're welcome! My pleasure. When $p=b$, we have $\cosh x+\sinh x = e^x$ and the integral $$I(a,b,b)=\frac12 \int_{-\infty}^{\infty}\sin(pa\cosh x+ab \sinh x)dx=\frac12\int_0^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{4}$$Thus, the value is average of the left-side and right-side limits. This is caused from the transition of a function going from even to odd. – Mark Viola Jul 28 '15 at 15:46
  • @Dr.MV Is there something that can be said about the original integral, pre-substitution? I mean that we can set $b = 1$ by doing substitution, so that the same singular behavior can be seen in \begin{equation} \int_0^{\infty} \frac{\sin(p \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}} \cos x dx \end{equation}. Is there something special that can be seen in this integral, without substitutions, at $p = 1$, or to expect some sort of transition at $p = 1$? – questions Jul 28 '15 at 15:52
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    @questions While with experience one might be able to "see" a discontinuity at $p=1$, I do not have the insight or foresight to deduce this by inspection alone. And without analysis, including numerical/empirical, the "magical" change in the value of the integral's going from $0$ at $p=.99999$ to $\approx. \pi/2$ at $p=1.000001$ does not appear obvious to me. – Mark Viola Jul 28 '15 at 16:33
  • @Mark Viola : Why $I(a,b,p)= \begin{cases} \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{p^2-b^2}\cosh x\right),dx&,0<b<p\\ \frac12\int_{-\infty}^{\infty}\sin\left(a\sqrt{b^2-p^2}\sinh x\right),dx&,b>p>0 \end{cases}$ ? Where did $\operatorname{arctanh}(A/B)$ (or $\operatorname{arctanh}(B/A)$) disappear to? – Plantation Mar 01 '23 at 07:10
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    @Plantation The terms are just translations in the hyperbolic cosine. Just make a simple substitution and they "disappear" since the integration is over the entire set of reals. – Mark Viola Mar 01 '23 at 14:43
  • @Mark Viola : Thanks! I have a feeling. – Plantation Mar 03 '23 at 03:09