A general integral equality $\int_{0}^{\infty}e^{-sx}\frac{\mathrm{d}x}{\left(1+x^{2}\right)^{a/2}}\left[s+\frac{ax}{1+x^{2}}\right]=1$

Question

How can we prove this integral equality:

$$\int_{0}^{\infty}e^{-sx}\frac{\mathrm{d}x}{\left(1+x^{2}\right)^{a/2}}\left[s+\frac{ax}{1+x^{2}}\right]=1 \tag{1}$$

I found it in a very roundabout way, but numerical experiments indicate that it's true for any values of $s,a \in \mathbb{C}$, provided $\Re (s) > 0$.

I found it by considering a family of integrals:

$$I_n (s)=\int_{0}^{\infty}e^{-sx}\frac{\mathrm{d}x}{\left(1+x^{2}\right)^{n+1/2}}$$

For which we can derive several recurrence relations, including:

$$I_{n}^{\prime\prime}(s)+I_{n}(s)=I_{n-1}(s) \tag{2}$$

And a 2nd order differential equation:

$$sI_{n}^{\prime\prime}(s)-\left(2n-1\right)I_{n}^{\prime}(s)+sI_{n}(s)=1 \tag{3}$$

My derivation of (3) is rather lengthy and involves several instances of integration by parts.

From (2) and (3) a particular case of (1) directly follows.

But I would like a better and more general proof of (1), as it seems rather interesting to me. Perhaps there are other known cases of such identities?

Answer 1

The integrand is exactly the derivative of $-\frac{e^{-sx}}{\left(1+x^2\right)^{\frac{a}{2}}}$, so evaluating at the endpoints gives the desired result.

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For a less hand-wavy solution, using integration by parts we get \begin{align*} I & = \int_{0}^{\infty}\color{purple}{se^{-sx}}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\mathrm{d}x+\int_{0}^{\infty}e^{-sx}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\frac{ax}{1+x^{2}}\mathrm{d}x \\ &= \int_{0}^{\infty}\color{purple}{\left( -e^{sx}\right)'}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\mathrm{d}x+\int_{0}^{\infty}e^{-sx}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\frac{ax}{1+x^{2}}\mathrm{d}x \\ & \overset{\text{IBP}}{=} -\frac{e^{-sx}}{\left(1+x^{2}\right)^{\frac{a}{2}}}\Bigg\vert_{0}^{\infty} +\int_{0}^{\infty}e^{sx}\left(\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\right)'\mathrm{d}x+\int_{0}^{\infty}e^{-sx}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\frac{ax}{1+x^{2}}\mathrm{d}x\\ & = 0 -(-1) - \int_{0}^{\infty}e^{-sx}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\frac{ax}{1+x^{2}}\mathrm{d}x + \int_{0}^{\infty}e^{-sx}\frac{1}{\left(1+x^{2}\right)^{\frac{a}{2}}}\frac{ax}{1+x^{2}}\mathrm{d}x\\ & =1 \end{align*}

where the condition $\Re\{s\} >0$ is used to argue $e^{-sx} \to 0$ as $x$ goes to infinity.