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I try to calculate the integral \begin{align} f(y,z):=\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx \end{align} on the set $y>0$ and $0<z<1$. I could show by some calculation that in the limit $z\to0$ holds \begin{align} f(y,0)=e^{y/2}K_0(y/2), \end{align} where $K_0$ denotes the modified Bessel function of the second kind and \begin{align} \lim_{z\to1}f(y,z)=f(y,1)=\infty,\\ \lim_{y\to0}f(y,z)=f(0,z)=\infty. \end{align} But I can't evaluate the integral in the interior of the defined set. Any help, ideas or hints how to solve this would be appreciated.


Here is a selection of what I've tried:

  • several substitutions (e.g. $\sinh$,$\dots$),
  • integration by parts,
  • tried to evaluate the integral $$\partial_zf(y,z)=\int_{\mathbb R} -yx\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx, $$
  • tried to evaluate the integrals $$\partial_yf(y,z)=\int_{\mathbb R} -(x^2+zx\sqrt{x^2+1})\cdot \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx $$ and $$f(y,z)-\partial_yf(y,z)=\int_{\mathbb R} (\sqrt{x^2+1}+zx)\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx. $$
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    $\begingroup$ A different form of the integral is $$ f(y,z) = \frac{{e^{y/2} }}{2}\int_{ - \infty }^{ + \infty } {\exp \left( { - \frac{y}{2}(\cosh s + z\sinh s)} \right)ds} . $$ I was not able to express it in terms of known functions (in general). $\endgroup$
    – Gary
    Commented Feb 5, 2021 at 13:42
  • $\begingroup$ This is exactly the way I used to represent $f(y,0)$ by the Bessel function. Maybe this form of the integral helps others to find a solution. I should have written it in the question above to save work. Thank you for your efforts. $\endgroup$
    – mag
    Commented Feb 5, 2021 at 16:04
  • $\begingroup$ It seems to be a weird mixture of a modified Bessel function and and Anger$-$Weber function (dlmf.nist.gov/11.10.E4). $\endgroup$
    – Gary
    Commented Feb 5, 2021 at 16:19

2 Answers 2

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\begin{align} f(y,z)=&\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx\\ =&\frac{1}{2}\int_{\mathbb R} \exp\left(-y(\sinh^2\frac{u}{2}+z\cdot \sinh\frac{u}{2}\cosh\frac{u}{2} ) \right)\mathrm du\\ =&\frac{1}{2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u -1+z\cdot \sinh u ) \right)\mathrm du\\ =&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u+z\cdot \sinh u ) \right)\mathrm du \end{align} Until this point I used the substitution $x=\sinh\frac{u}{2}$ and the equation $\sinh^2\frac{u}{2}=\frac{1}{2}(\cosh u -1)$, as well as $\sinh\frac{u}{2}\cosh\frac{u}{2}=\frac{1}{2}\sinh u$. From this I could see for the case $z=0$, that we have $f(y,0)=e^{y/2}K_0(y/2)$ using $$K_0(\phi)=\int_0^\infty\exp\left(-\phi\cosh x \right)\mathrm dx,\qquad \forall \phi>0.$$ To go on from this someone helped me with the hint $$ \cosh u+z\cdot \sinh u =\sqrt{1-z^2}\cosh(u+\tanh^{-1} z).$$ This brings us to \begin{align} f(y,z)=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh(u+\tanh^{-1} z) \right)\mathrm du\\ =&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\ =&e^{y/2}\int_{0}^\infty \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\ =&e^{y/2}K_0\left(\frac{y}{2}\sqrt{1-z^2}\right). \end{align}


The idea behind the hint was to find values $\beta, v$ such that we can use the addition theorem \begin{align} \cosh u+z\cdot \sinh u=\frac{1}{\beta}(\beta\cosh u+z\beta\cdot \sinh u)\overset{!}{=}&\frac{1}{\beta}(\cosh v\cosh u+\sinh v\sinh u)\\ =&\frac{1}{\beta}\cosh(u+v), \end{align} as a shift $v$ of the argument doesn't affect the integral since we are integrating from $-\infty$ to $\infty$. To find these values we need to have $\beta=\cosh v$ and $z\beta=\sinh v$. This brings us to $$v=\tanh^{-1}z$$ and $$\beta=\frac{1}{\sqrt{1-z^2}}.$$

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not an answer

I computed numerically the integral in the interval $[0,10]\times[0,1]$ $$F(y,z)=\int_{-\infty }^{\infty } \frac{\exp \left(-y \left(xz\sqrt{x^2+1} +x^2\right)\right)}{\sqrt{x^2+1}} \, dx$$ maybe it can be useful

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