# Showing two integrals are equal

I would like to show that

$$\frac{1}{\Gamma\left(\, -\alpha\,\right)} \int_{-\infty}^{x}\frac{\,\mathrm{f}\left(\, t\,\right)} {\left(\, x - t\,\right)^{\alpha + 1}}\,\mathrm{d}t = \frac{1}{2\pi} \int_{-\infty}^{\infty}(\mathrm{i}t)^{\alpha}\,\mathrm{e}^{\mathrm{i}kt} \int_{-\infty}^{\infty}\,\mathrm{e}^{-\mathrm{i}kt} \,\mathrm{f}\left(\, k\,\right)\,\mathrm{d}k\,\mathrm{d}t$$

for fixed $\alpha$ such that $\Re(\alpha) < 0$ and square integrable $f$. I was under the impression that this is a straight forward computation, but I am having difficulty getting anywhere.

I apologize for the weak effort in advance.

So far, I've tried considering rewriting the LHS in terms of the Cauchy Integral formula by writing something like

$$\frac{1}{\Gamma(-\alpha)} \left(\int_{-\infty}^{\infty} \frac{f(t)}{(x-t)^{\alpha+1}}dt - \int_{x}^{\infty} \frac{f(t)}{(x-t)^{\alpha+1}}dt\right)$$

but I can't get anything that is both useful and makes sense from here. I've also considered trying to solve one of the integrals on the RHS using some contour integration technique by considering something like

$$\int_{-\infty}^{\infty} e^{-itx}f(x)dx = \lim_{R\rightarrow \infty}\int_{-R}^{R} e^{-itz} f(z)dz + \int_{\gamma} e^{itz}f(z)dz$$

but it is difficult for me to decide how I can choose a $\gamma$ which makes sense knowing nothing about the poles of $f$.

The only other ideas I have might be to try to use integration by parts to rewrite both sides using properties of the Fourier transform or possibly try to rewrite the integrals as infinite sums but I'm not getting anywhere.

• Are you sure the equality holds? The LHS seems to depend on $x$, while the RHS doesn't... Jul 28, 2016 at 12:52
• @PierpaoloVivo The inner integral on the RHS integrates out the $x$ and becomes a function of $t$. The outer integral still contains an $e^{itx}$, but integrates out the $t$ leaving it depending on $x$. Jul 28, 2016 at 12:55
• Ok, then I wouldn't use $x$ as integration variable for the inner integral on the RHS. Use $y$ instead (the way it's written is confusing). Jul 28, 2016 at 12:57
• @PierpaoloVivo Fair enough, I thought it would make it clearer that the RHS can be written as $\mathscr{F}^{-1}((it)^{\alpha}\mathscr{F}(f(x)))$ like this. Jul 28, 2016 at 13:00
• Ok, got it. Also, some conditions are needed on $\alpha$ as well, aren't they? If $\alpha$ is "too small", the LHS may not converge... Jul 28, 2016 at 13:04

## 1 Answer

Just insert the identity $$(\mathrm{i} t)^{-k}=\frac{1}{\Gamma(k)}\int_0^\infty \omega^{k-1}e^{-\mathrm{i}\omega t}d\omega\ ,$$ in the RHS, with $k=-\alpha$. This gives $$\frac{1}{2\pi\Gamma(-\alpha)} \int_0^\infty d\omega\ \omega^{-\alpha-1} \int_{-\infty}^{\infty} f(y)dy\int_{-\infty}^{\infty}dt\ e^{it(x-\omega-y)}\ ,$$ and you can recognise the integral representation of the Dirac delta on the far right, $$\int_{-\infty}^\infty \frac{dt}{2\pi}e^{\mathrm{i} t s}=\delta(s)\ .$$ Therefore your RHS is equal to $$\frac{1}{\Gamma(-\alpha)} \int_0^\infty d\omega\ \omega^{-\alpha-1} \int_{-\infty}^{\infty} dy\ f(y)\delta(x-\omega-y)\ ,$$ and you can now use the Dirac delta to kill the omega-integral, by imposing that $\omega=x-y>0$ (as the range of the $\omega$-integral is only $(0,\infty)$).

In the end, you get $$\frac{1}{\Gamma(-\alpha)}\int_{-\infty}^{x} dy\ (x-y)^{-\alpha-1} f(y)\ ,$$ which is your LHS as desired.

• This is wonderful. I did not expect the Gamma distribution to show up like that! Thank you very much. Jul 28, 2016 at 14:55