Integral $ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$ I am trying to solve the following definite integral:
$$ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{(x-a)^2 + b^2}\,\,t)}{(x-a)^2 + b^2} dx$$
with $a$ and $b$  being real constants. Notice that the integrand is non-negative for all x , with a peak around $x=a$ for small values of $b/a$.  So the integration does give a finite value. 
To go about this,  I tried substituting $(x-a)^2 + b^2 = y^2$
This gives me $dx \,(x - a) = y \, dy$  whence:
$$ \int_{?}^{+\infty} \frac{ \sin^2(|y|\,\,t)}{y} \frac{1}{\sqrt{y^2 - b^2}} dy$$
While the upper limit of y transforms to $+\infty$, clearly the lower limit is not $-\infty$ (even if it is $-\infty$ the integrand is odd so it evaluates to zero which cannot be true).
This suggests its a case of bad substitution. The only better alternative I can think of is to substitute $x-a = y$, this gives me:
$$ \int_{-\infty}^{+\infty} \frac{ \sin^2(\sqrt{y^2 + b^2}\,\,t)}{y^2 + b^2}  dy $$
Any thoughts on how I can get an expression for this integral ?
Thanks for your time!
 A: I assume $t, b> 0$. Note that your integral does not change when $b$ and/or $t$ is replaced by its negative. The case $t=0$ is trivial and the case $b=0$ follows by continuity. After a change of integration variables ($w = \frac{{x - a}}{b}$), your integral becomes
$$
\frac{2}{b}\int_0^{ + \infty } {\frac{{\sin ^2 (b\sqrt {w^2  + 1} t)}}{{w^2  + 1}}dw} .
$$
Differentiation with respect to $t$ gives
$$
\frac{d}{{dt}}\frac{2}{b}\int_0^{ + \infty } {\frac{{\sin ^2 (b\sqrt {w^2  + 1} t)}}{{w^2  + 1}}dw}  = 2\int_0^{ + \infty } {\frac{{\sin (2b\sqrt {w^2  + 1} t)}}{{\sqrt {w^2  + 1} }}dw} \\ = 2\int_1^{ + \infty } {\frac{{\sin (2but)}}{{\sqrt {u^2  - 1} }}du}  = \pi J_0 (2bt),
$$
where $J_0$ is the Bessel function of the first kind of order zero (cf. http://dlmf.nist.gov/10.9.E12). Since your integral vanishes at $0$, it must equal to
$$
\pi \int_0^t {J_0 (2bs)ds} .
$$
This can be expressed in terms of Bessel and Struve functions if you like. Note that by replacing $t$ and $b$ in the above by $|t|$ and $|b|$, it yields the formula for all real $t$ and $b$.
A: Thanks Gary. Your solution was quite impressive. 
Let me complicate the integral to a level higher and see if we can proceed similarly to get an analytical solution.  I now write the integral to be solved as:
$$ \int_{-\infty}^{+\infty} \frac{\sin^2 \sqrt{(x^2-a^2)^2 + b^4  }\, t}{(x^2-a^2)^2 + b^4 } \, \, dx $$
You will observe that for $x \approx a$ we can write $x^2 - a^2 \approx 2 a (x - a)^2 $ thus reducing this integral to a form similar to what you have solved previously.  However, I now want to arrive at a solution without making such an approximation.
I proceed now similar to what you had done previously.  
I attempted to re-write the integral by replacing $\left(\frac{x^2- a^2}{b^2} \right)^2 = w^2 $ - however the even integrand will no longer continue to be even with this substitution. So I continue without such a replacement. 
$$ \frac{2}{b^4} \int_{0}^{+\infty} \frac{\sin^2 b^2 t \sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1}}{ \left(\frac{x^2-a^2}{b^2} \right)^2 +1  } \,\,dx $$
Differentiating with respect to time, this can be shown to reduce to:
$$ \frac{2}{b^2} \int_{0}^{+\infty} \frac{\sin 2 b^2 t \sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1}}{ \sqrt{ \left(\frac{x^2-a^2}{b^2} \right)^2 +1 } } \,\,dx $$
Now comes the hard part:  I attempt replacing $\sqrt{\left(\frac{x^2-a^2}{b^2} \right)^2 +1 } = u $.  This integral will be equivalent to:
$$2 b^2 \int_{\sqrt{\frac{a^4}{b^4}+1}}^{+\infty} \frac{\sin 2 b^2 u t }{ \sqrt{u^2 - 1} } \,\, \frac{1 }{\sqrt{a^2 + b^2 \sqrt{u^2 - 1}  }} du $$ At this point I have no idea how to proceed. Any ideas ?
Thanks!
