A convolution like integral equation

I would like to solve the following integral equation for $g(z)$. $$\int_z^\infty g(\zeta)(\zeta-z)^{\alpha-1} d\zeta = e^{-bz}, \tag{1}$$ where $\alpha$ and $b$ are constants.

I would also like to know the techniques for solving $g$ in the more general equation $$\int_z^\infty g(\zeta)h(\zeta-z) d\zeta = f(z), \tag{2}$$ for known $f$ and $h$.

We can put the second equation in the alternative form $$\int_0^\infty g(z(x+1))h(x) dx = f(z). \tag{3}$$ I do not know if it helps though.

I put equation $(2)$ into the form $$\int_{-\infty}^\infty g(\zeta)\Theta(\zeta-z)h(\zeta-z) d\zeta = f(z),$$ where $\Theta$ is the Heaviside step function, which suggests using Fourier transform applied to convolution. However, for the kind of $h$ and $f$ as in equation $(1)$, the Fourier transform $F[g](k) = \int_{-\infty}^\infty g(x)e^{ikx}dx$ is divergent thus not readily defined. Perhaps I need to get $k$ into the complex plane to make the transform sensible for those kinds of functions.

I have tried the Laplace transform which is a special case of Fourier transform. It does not appear to work because of the limits of the integration. Maybe someone can transform the form of the integral to make it work.

Perhaps there is a more direct method. Any ideas?

• Here is something to try. Rewrite (1) as $$(1-p)G(z) - \mathrm{e}^{-\beta z} = p\int_z^{\infty} G(\zeta) (\zeta - z)^{\alpha} \ \mathrm{d}\zeta,$$ Now let $G(z,p) = G_0 +pG_1+p^2G_2 + \ldots$ and $g(z) = \displaystyle{ \lim_{p \rightarrow 1}} \ G(z,p)$ Matching coefficients of powers of $p$ gives \begin{align} p^0:& \qquad G_0 = -\mathrm{e}^{-\beta z}, \\ p^1:& \qquad G_1 + \mathrm{e}^{-\beta z} = \int_z^{\infty} G_0(\zeta) (\zeta - z)^{\alpha} \ \mathrm{d}\zeta, \end{align} making the substitution here of $\beta u=\zeta - z$, we arrive at a Gamma function type integral. – Bennett Gardiner Jan 6 '14 at 5:38
• These ideas come from this paper. Convergence issues will be up to you to check. – Bennett Gardiner Jan 6 '14 at 5:39
• Try the Laplace transform instead of the Fourier transform. I find that they tend to work well when the transforms integrate on only one side of the variable. And it is equivalent to what you were thinking about evaluating the Fourier transform along, say, the imaginary axis. – Stephen Montgomery-Smith Jan 6 '14 at 6:28
• The integral is a Weyl’s fractional integral. If it’s allowed to use the famous Harry Batmann’s “Tables of Integral Transforms”, Vol.2, page 202, line 11, (MacGraw-Hill edit. 1954) we see that the function g is the inverse transform of exp(-bz) which also is the exponential function. That is a short cut to the result : g(z)=(b^(a+1)) exp(-bz)/Gamma(a+1) – JJacquelin Jan 6 '14 at 16:03
• @BennettGardiner: Thank you for the idea and reference. I think this series appears to converge to the fractional integral solution below, but I have not checked yet. Please feel free to comment if you have seen the agreement already. – Hans Jan 6 '14 at 21:48

Consider $$\int_z^\infty g(\zeta)(\zeta-z)^{\alpha-1} d\zeta = f(z), \tag{1}$$ where Re$(\alpha)>0$.

For now we assume $f(z)=0$, for $z<0$.

Rewrite equation $(1)$ as $$\int_{-\infty}^\infty g(\zeta)(\zeta-z)_+^{\alpha-1} d\zeta = f(z), \tag{1.1}$$

Fourier transform equation $(1.1)$. $$F[x_+^{\alpha-1}](-k)F[g](k) = F[f](k), \tag{2}$$ where $F[f](k)$ stands for the Fourier transform in complex variable $k$ of function $f$. The difficulty mentioned in the question regarding the convergence of Fourier transform of functions like the exponential function can be circumvented either by limiting first the support of the function to, say, the positive axis, or expand the function space to Gelfand-Shilov space, or simply forge ahead without worrying about the convergence at all.

$$F[x_+^{\alpha-1}](-k) = \int_0^\infty e^{-ikx}x^{\alpha-1} dx = (ik)^{-\alpha}\Gamma(\alpha), \,\mbox{Im}(k)<0, \tag{3}$$ where the contour is deformed to the real axis.

$$F^{-1}\Big[\frac{(ik)^\alpha}{\Gamma(\alpha)}\Big](x) = \frac{1}{\Gamma(\alpha)\Gamma(-\alpha)}(-x)_+^{-\alpha-1}. \tag{4}$$ Therefore, $g$ would be the convolution of the product of $(4)$ and $f$, or

$$g(x) = \frac{1}{\Gamma(\alpha)\Gamma(-\alpha)}\int_0^\infty y^{-\alpha-1}f(x+y)dy$$ (to be continued, as the left limit does not converge, and thus there should be another term to cancel the singularity.)

The following is a direct method that is inspired by the above Fourier transform.

Integrate both side with respect to $dz(x-z)^{\beta-1}$, Re$(\beta)>0$ $$\int_x^\infty dz (z-x)^{\beta-1}\int_z^\infty g(\zeta)(\zeta-z)^{\alpha-1} d\zeta =\int_x^\infty d\zeta g(\zeta)\int_x^\zeta dz (z-x)^{\beta-1}(\zeta-z)^{\alpha-1}=\int_x^\infty dz (z-x)^{\beta-1}f(z),$$ where the first equality comes from changing order of integration (by checking the conditions of Fubini's theorem which may require more regularity conditions on $g$ which I am going to do right now). $$\int_x^\zeta (z-x)^{\beta-1}(\zeta-z)^{\alpha-1}dz = (\zeta-x)^{\alpha+\beta-1}\int_0^1 t^{\beta}(1-t)^\alpha dt=(\zeta-x)^{\alpha+\beta-1} B(\beta,\alpha),$$ where $B$ is the Beta function. So by setting $\beta=1-\alpha$, we shall have $$B(1-\alpha,\alpha)\int_x^\infty d\zeta g(\zeta) = \int_0^\infty z^{-\alpha}f(x+z)dz,$$ and $$g(x) = -\frac{1}{B(1-\alpha,\alpha)}\int_0^\infty z^{-\alpha}f'(x+z)dz. \tag{2}$$

For the original problem $f(z) = e^{-bz}$, equation $(2)$ becomes $$g(x) = \frac{b^\alpha e^{-bx}}{\Gamma(\alpha)},$$ where $\Gamma$ is the Gamma function.

Therefore, it turns out $e^{-bz}$ is an eigenfunction of the integral operator $\int_z^\infty d\zeta (\zeta-z)^{\alpha-1}g(\zeta)$ on function $g(z)$.

• I think you need to check your last step. The exponential inside the integral will still have a $b$ term in it? – Bennett Gardiner Jan 7 '14 at 1:31
• @BennettGardiner: Oops, typo. Corrected the sign in equation $(2)$ and exponent on $b$ in the last line now. Thanks. – Hans Jan 7 '14 at 2:11
• I was thinking in an Abel Integral Equation "trick" when I saw your amazing solution. – Felix Marin Jan 7 '14 at 3:03
• @FelixMarin: The Abel integral equation looks like a special case of this integral equation. Thank you. :-) – Hans Jan 7 '14 at 17:49

Rewrite (1) as $$(1-p)G(z) + p\int_z^{\infty} G(\zeta) (\zeta - z)^{\alpha-1} \ \mathrm{d}\zeta = \mathrm{e}^{-\beta z},$$ Now let $G(z,p) = G_0 +pG_1+p^2G_2 + \ldots$ and $g(z) = \displaystyle{ \lim_{p \rightarrow 1}} \ G(z,p).$

Matching coefficients of powers of $p$ gives \begin{align} p^0:& \qquad G_0 = \mathrm{e}^{-\beta z}, \\ p^1:& \qquad G_1 - G_0 = -\int_z^{\infty} G_0(\zeta) (\zeta - z)^{\alpha-1} \ \mathrm{d}\zeta, \end{align} making the substitution here of $u=\beta(\zeta - z)$, we arrive at \begin{align} G_1 - \mathrm{e}^{-\beta z} &= -\frac{\mathrm{e}^{-\beta z}}{\beta^{\alpha}} \int_0^{\infty} \mathrm{e}^{- u}u^{\alpha-1} \ \mathrm{d}\zeta,\\ G_1 &= \mathrm{e}^{-\beta z}\left( 1-\frac{\Gamma(\alpha)}{\beta^\alpha}\right). \end{align} Note that this is a constant multiple of $G_0$, which suggests $$G_n = \mathrm{e}^{-\beta z}\left(1-\frac{\Gamma(\alpha)}{\beta^\alpha}\right)^n$$ Summing the series and letting $p \rightarrow 1$ gives us a geometric series where $$g(z) = \frac{\mathrm{e}^{-\beta z}}{1-\left(1-\frac{\Gamma(\alpha)}{\beta^\alpha}\right)},$$ or $$g(z) = \frac{\beta^{\alpha}\mathrm{e}^{-\beta z}}{\Gamma(\alpha)},$$ which agrees with the other method.

We have the restriction that $$-1<\frac{\Gamma(\alpha)}{\beta^\alpha}-1 < 1$$ for convergence.

(The following sentence no longer applies and convergence domain is a subset of what is expected, as the homotopy operator has been modified in light of a comment below.) This is a rather strange condition, it implies that $\lfloor\alpha\rfloor$ is a negative odd integer, according to the graph of the gamma function.

• Check out my second solution. For your method, I think you can also adjust the range of parameters, e.g., $\alpha$ and $b$, for reason of convergence by adjusting the form of homotopy function. – Hans Jan 7 '14 at 4:17
• Yeah, cool. I imagine it's an art as to how to construct the function. What do your calculations say about convergence, then? Is this a valid solution for all $\left\{\alpha, b \right\}$? (Or $b>0$, $\alpha$ not a negative integer?) – Bennett Gardiner Jan 7 '14 at 4:33
• You may have made a mistake in setting up the homotopy operator. The form you have chosen, though homotopic, is a bit strange as it is not convex. It is fine by itself. However, the resulting convergence condition forces $\alpha$ to be negative. But this forces both integration on exponential functions nonconvergent. You can salvage the situation by moving $(1-p)G$ to the right side of the homotopy operator equation. – Hans Jan 7 '14 at 18:04
• Regarding your question about the convergence in my second solution, choosing large enough positive $c$ makes all positive $\alpha$ viable. – Hans Jan 7 '14 at 18:07
• Awesome - nice work! – Bennett Gardiner Jan 7 '14 at 22:38

Here is another solution.

Let $S$ be an operator on some function space yet to be prescribed such that $Sh(z)=\int_z^\infty d\zeta (\zeta-z)^{\alpha-1}h(\zeta)$ on a member function $h$ of that space. Assume $S$ is invertible. $$Sg = f$$ $$\implies$$ $$g=S^{-1}f=(C-(C-S))^{-1}f=(I-(I-C^{-1}S))^{-1}C^{-1}f=\sum_{i=0}^\infty (I-C^{-1}S)^i C^{-1}f, \tag{1}$$ for some invertible operator $C$ and if $\|(I-C^{-1}S)^i\|<1$ for the prescribed function space,

Now for this particular function space $\{f(z)=e^{-bz}|\mbox{Re}(b)>0\}$, $Sf = b^{-\alpha}\Gamma(\alpha)f$ a scalar multiplication, or in other words, the exponential function space is an eigenfunction space of $S$. Pick, say, an arbitrary multiplication operator $C=c$ for a constant $c$ (positive or negative), $I-C^{-1}S = 1-c^{-1}b^{-\alpha}\Gamma(\alpha)$. So long as $0<c^{-1}b^{-\alpha}\Gamma(\alpha)<2$, equation $(1)$ converges and to $S^{-1}$, which in this eigenfunction space is simply the reciprocal of the original eigenvalue, or $$g(z) = S^{-1}e^{-bz}=\frac{b^\alpha}{\Gamma(\alpha)}e^{-bz}.$$

• At first sight, your result: (b^a)*exp(-bz)/gamma(a) is not the same as mine : (b^(a+1))*exp(-bz)/Gamma(a+1). This is probably due to the exponent which was changed from (a) to (a-1). Apart from that I agree. – JJacquelin Jan 7 '14 at 8:10
• @JJacquelin: Sorry, exactly as you supposed, this is due to I having changed $\alpha$ to $\alpha-1$ for presentational aesthetics. – Hans Jan 7 '14 at 18:09