Cfrgtkky

I'd really love to evaluate this integral exactly in terms of known functions, because for large $b$ it becomes a pain numerically.

$$I(b,s)=int_0^1 frac{cos bx}{sqrt{x^2+s^2} }dx$$

Didn't get anywhere wth integration by parts, or series. I want an expression valid for both large and small values of the parameters.

Note: we have $b= pi n$, where $n$ is an integer.

For $s gg 1$ we can expand the root as a series and get a good approximation. However, this case is of small use to me, as in general $s$ is of the order of $1$ or smaller.

So here's my latest attempt:

Edited

$$x=s sinh v$$

$$I(b,s)=int_0^{sinh^{-1} frac{1}{s}} cos left(bs sinh v right) dv$$

Let's try integration by parts:

$$U=cos left(bs sinh v right) \ dV=dv$$

$$dU=-bssin left(bs sinh v right) cosh v \ V=v$$

$$I(b,s)=cos b sinh^{-1} frac{1}{s}+bs int_0^{sinh^{-1} frac{1}{s}} v sin left(bs sinh v right) cosh v dv$$

$$I(b,s)=cos b sinh^{-1} frac{1}{s}+bs int_0^{frac{1}{s}} sinh^{-1} r sin left(bs r right) dr$$

$$I(b,s)=cos b sinh^{-1} frac{1}{s}+b int_0^1 sinh^{-1} frac{x}{s} sin left(b x right) dx$$

This seems a little better, at least we separated the special case $I(0,s)$, which is the first term. Not sure how to continue.

Using the fact that $b= pi n$, we can transform the integral as:

$$b int_0^1 sinh^{-1} frac{x}{s} sin left(b x right) dx=pi sum_{k=0}^{n-1} (-1)^k int_0^1 sinh^{-1} left( frac{t+k}{ns} right) sin pi t dt$$

The function $sinh^{-1}$ is pretty nice, it has a logarithmic growth, and doesn't blow up anywhere. I guess, this could be the way to get a good approximation, especially since we have an alternating sum.

Getting a grasp of the original problem: we want to find the asymptotic behaviour of the Fourier coefficients of $frac{1}{sqrt{x^2+s^2}}in L^2(0,1)$. Since the inverse Laplace transform of $frac{1}{sqrt{x^2+s^2}}$ is $J_0(as)$ and the Laplace transform of $cos(pi n x)mathbb{1}_{(0,1)}(x)$ is $frac{a}{a^2+n^2pi^2}-frac{a(-1)^n}{e^a(a^2+pi^2 n^2)}$, we have

$$ int_{0}^{1}frac{cos(pi n x)}{sqrt{x^2+s^2}},dx = underbrace{int_{0}^{+infty}frac{a J_0(as)}{a^2+pi^2 n^2},da}_{K_0(pi n s)} + (-1)^{n+1}int_{0}^{+infty}frac{a J_0(as)}{e^a(a^2+pi^2 n^2)},da $$
which is fairly easy to approximate numerically due to the known bounds for Bessel functions.

Close to the origin $J_0(a)$ can be approximated through its rapidly-convergent Maclaurin series, far from the origin we have Tricomi's $J_0(a)approxfrac{sin(a)+cos(a)}{sqrt{pi a}}$. For $K_0$ we have Hankel's expansion.

In particular the Fourier coefficients of $frac{1}{sqrt{x^2+s^2}}$ decay like $frac{1}{n^2 s^3}$.

It is also interesting to point out a curious inequality provided by Cauchy-Schwarz:

$$begin{eqnarray*}left|int_{0}^{+infty}frac{a J_0(as)}{e^a(a^2+pi^2 n^2)},daright|^2&leq& int_{0}^{+infty}frac{J_0(as)^2}{e^{2a}},daint_{0}^{+infty}frac{a^2}{(a^2+pi^2 n^2)^2},da\&=&frac{1}{8n,text{AGM}(1,sqrt{1+s^2})}leqfrac{1}{8n}(1+s^2)^{-1/4}.end{eqnarray*}$$

The asymptotic for large $n$ is determined by the behaviour of the integrand at $x = pm 1$. Choosing a contour going in the direction $i$ from $x = -1$ and then in the direction $-i$ towards $x = 1$ and expanding the non-exponential part, we have
$$frac 1 {sqrt {x^2 + s^2}} biggrvert_{x = -1 + i xi} =
frac 1 {sqrt {1 + s^2}} + frac {i xi} {(1 + s^2)^{3/2}} + O(xi^2), \
frac 1 {sqrt {x^2 + s^2}} biggrvert_{x = 1 + i xi} =
frac 1 {sqrt {1 + s^2}} - frac {i xi} {(1 + s^2)^{3/2}} + O(xi^2).$$
The contributions from the first terms will cancel out, leaving
$$I(pi n, s) =
frac 1 2int_{-1}^1 frac {e^{i pi n x}} {sqrt {x^2 + s^2}} dx sim
i int_0^infty
frac {i xi} {(1 + s^2)^{3/2}} e^{-i pi n - pi n xi} dxi = \
frac {(-1)^{n - 1}} {pi^2 (1 + s^2)^{3/2} n^2},
quad n to infty, ,n in mathbb N.$$

Just a comment

The integral is the real part of
$$int^1_0frac{e^{ibx}}{sqrt{s^2+x^2}}dx$$

If we make a substitution $x=iscos t$, without dealing with branch cuts rigorously we can obtain

$$-iint^{cos^{-1}(-i/s)}_{pi/2}e^{-bscos t}dt$$

Essentially,
$$I(b,s)=Im~ int^{cos^{-1}(-i/s)}_{pi/2}e^{-bscos t}dt $$
or
$$I(b,s)=-Im~ int^{sin^{-1}(-i/s)}_{0}e^{-bssin t}dt $$

which suggests a relation with Bessel functions. However, due to the upper and lower limits not being $pmpi$, the normal Bessel functions cannot be used. I think there might be a need to define some kind of ‘incomplete Bessel function’ to write the integral in closed form.