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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!

jimjim
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Karthik
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2 Answers2

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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 $b$.

Gary
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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!

Karthik
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