13
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ Differentiate with regard to $\beta$, let $x=\alpha\sinh t$, and use $\sin u=\Im(e^{iu})$. $\endgroup$ – Lucian Jun 2 '14 at 20:17
7
$\begingroup$

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 $$\begin{aligned} \sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) &=\sqrt{\frac{1-\cos\left(\arctan\frac{c}{b}\right)}{2}}\\ &= \frac{1}{\sqrt{2}}\sqrt{1-\frac{b}{\sqrt{b^2+c^2}}}\\ &= \frac{1}{\sqrt{2}}\sqrt{\frac{\sqrt{b^2+c^2}-b}{\sqrt{b^2+c^2}}}=\frac{1}{\sqrt{2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}} \end{aligned}$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I am looking for elementary methods (and I feel there is one), no residue calculus please. :) +1 though. $\endgroup$ – Pranav Arora Jun 11 '14 at 10:26
  • $\begingroup$ @PranavArora Thanks for putting this up for bounty my friend. $\endgroup$ – Jeff Faraci Jun 11 '14 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. :) $\endgroup$ – Pranav Arora Jun 11 '14 at 20:03
  • 1
    $\begingroup$ Integrals containing $\exp(\cosh x)$ are expressible in terms of Bessel functions. $\endgroup$ – Lucian Jun 15 '14 at 7:33
  • $\begingroup$ It looks like the equation for $F'(\beta)$ is missing an overall factor of $(-1)$. $\endgroup$ – David H Jun 16 '14 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.