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I am having trouble showing this equality is true$$ \int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx=\sqrt\frac{\pi}{2}\frac{\gamma \exp\big(-\alpha\sqrt{\gamma^2+\beta^2}\big)}{\sqrt{\beta^2+\gamma^2}\sqrt{\beta+\sqrt{\beta^2+\gamma^2}}}, $$ $$ \mathcal{Re}(\alpha,\beta,\gamma> 0). $$ I do not know how to approach it because of all the square root functions.

It seems if $x=\pm i\alpha \ $ we may have some convergence problems because of the denominator. Perhaps there are ways to solve this using complex methods involving the branch cut from the square root singularity. I just do not know what to choose $f(z)$ for a suitable complex function to represent the integrand.

I also tried differentiating under the integral signs w.r.t $\alpha,\beta,\gamma$ but it did not simplify anything. Thanks. How can we calculate this integral?

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    $\begingroup$ Differentiate with regard to $\beta$, let $x=\alpha\sinh t$, and use $\sin u=\Im(e^{iu})$. $\endgroup$
    – Lucian
    Commented Jun 2, 2014 at 20:17

2 Answers 2

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Replace $\alpha$, $\beta$ and $\gamma$ with $a$, $b$ and $c$ respectively.

With the substitution $x=a\sinh t$, the integral can be written as: $$\begin{aligned} I & = \sqrt{2a}\int_0^{\infty} e^{-ab\cosh t}\sin(ac\sinh t)\sinh \left(\frac{t}{2}\right)\,dt \\ &=-\sqrt{2a}\Im\left(\int_0^{\infty} e^{-ab\cosh t}\cos\left(ac\sinh t+\frac{it}{2}\right)\,dt \right) \end{aligned}$$

Thanks to sir O.L. for evaluating the final integral here: Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$

The result is hence, $$\begin{aligned} I & = -\sqrt{2a}\Im\left(e^{-\frac{i}{2}\arctan\frac{c}{b}}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\right) \\ &=\sqrt{2a}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}} \sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) \\ &=\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}} \\ &=\boxed{\sqrt{\dfrac{\pi}{2}}\dfrac{c\exp\left(-a\sqrt{b^2+c^2}\right)}{\sqrt{b^2+c^2}\sqrt{\sqrt{b^2+c^2}+b}}} \end{aligned}$$ I used the following to simplify the above expression $$\begin{aligned} \sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) &=\sqrt{\frac{1-\cos\left(\arctan\frac{c}{b}\right)}{2}}\\ &= \frac{1}{\sqrt{2}}\sqrt{1-\frac{b}{\sqrt{b^2+c^2}}}\\ &= \frac{1}{\sqrt{2}}\sqrt{\frac{\sqrt{b^2+c^2}-b}{\sqrt{b^2+c^2}}}=\frac{1}{\sqrt{2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}} \end{aligned}$$

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As Lucian said. Take $\alpha,\beta,\gamma>0$ reals (once you are done you can extend it analytically). $$ F(\beta):=\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx $$ \begin{eqnarray*} F^\prime(\beta)&=&\int_0^\infty {\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}\sin (\gamma x)\,dx \\ &=& \int_0^\infty \alpha^{3/2}\sqrt{2}\sinh\left(\frac{t}{2}\right)\,\exp\big({-\beta\alpha\cosh(t)\big)}\sin (\gamma\alpha \sinh(t))\cosh(t)\,dt \end{eqnarray*} Now you have a nice analytic integrand, you can residue formula it away.

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    $\begingroup$ I am looking for elementary methods (and I feel there is one), no residue calculus please. :) +1 though. $\endgroup$ Commented Jun 11, 2014 at 10:26
  • $\begingroup$ @PranavArora Thanks for putting this up for bounty my friend. $\endgroup$ Commented Jun 11, 2014 at 19:57
  • $\begingroup$ @Integrals: I was working on this problem for a week but I couldn't come up with anything useful so I decided to put a bounty. :) $\endgroup$ Commented Jun 11, 2014 at 20:03
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    $\begingroup$ Integrals containing $\exp(\cosh x)$ are expressible in terms of Bessel functions. $\endgroup$
    – Lucian
    Commented Jun 15, 2014 at 7:33
  • $\begingroup$ It looks like the equation for $F'(\beta)$ is missing an overall factor of $(-1)$. $\endgroup$
    – David H
    Commented Jun 16, 2014 at 20:04

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