Return to Answer

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does indicate why a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$).

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

As a further check, note that the derivative with respect to $b$ gives formula 3.876.2 of Gradshteyn.

_{The corresponding formulas in Erdelyi contain several typo's. These have been corrected in Gradshteyn, the formulas in question have a label $^6$.}

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does indicate why a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$).

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

Notes:
_{• As a further check, note that the derivative with respect to $b$ gives formula 3.876.2 of Gradshteyn.
• The corresponding formulas in Erdelyi contain several typo's. These have been corrected in Gradshteyn, the formulas in question have a label $^6$.
• The integral also implies the identity $$\int_0^1 K_0\left(\sqrt{1-t^2}\right)\,dt=\tfrac{1}{2e} \text{Ei}(1)-\tfrac{e}{2} \text{Ei}(-1).$$}

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does suggest that a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$).

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

The discontinuous derivative at $y=b$ is present in all Fourier transforms of this type [sine or cosine of $b\sqrt{a^2+x^2}$ times some power of $(a^2+x^2)$].

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does indicate why a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$).

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

As a further check, note that the derivative with respect to $b$ gives formula 3.876.2 of Gradshteyn.

_{The corresponding formulas in Erdelyi contain several typo's. These have been corrected in Gradshteyn, the formulas in question have a label $^6$.}

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does suggest that a fully closed-form expression will not be forthcoming.

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

This is an integral in Oberhettinger's Table of Fourier Transforms (page 27, with $r=\sqrt{a^2+x^2}$). It is not a very helpful result, but it does suggest that a fully closed-form expression will not be forthcoming (you would need the indefinite integral of a Bessel function of argument $\sqrt{1-x^2}$).

There is a typo in the formula, a numerical check does not match. For $y<b$ a square root is missing in the argument of the Bessel $Y_0$ function, the following expressions do pass a numerical check:

$$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\frac{e^{-a y} \,\text{Ei}(a y)-e^ {ay} \,\text{Ei}(-a y)}{2 a}$$ $$\qquad\qquad-\frac{\pi}{2} \int_y^b Y_0\left(a \sqrt{t^2-y^2}\right) \, dt,\;\;0\leq y<b,$$ $$\int_{0}^\infty \frac{\sin\left(b\sqrt{a^2+x^2}\right)}{a^2+x^2}\cos yx\,dx =\int_0^b K_0\left(a \sqrt{y^2-t^2}\right) \, dt,\;\;y>b.$$

The discontinuous derivative at $y=b$ is present in all Fourier transforms of this type [sine or cosine of $b\sqrt{a^2+x^2}$ times some power of $(a^2+x^2)$].