$$ \mbox{How can I compute}\ \int_{0}^{\infty}\frac{\cos\left(ax\right)}{\left(1 + x^{2}\right)\,\sqrt{\,{x}\,}}\,\mathrm{d}x\ ?. $$
I saw many questions about computations of that integral without the term $\,\sqrt{\,{x}\,}\,$ and that it can be solved by using various techniques ( contour integral, iterated integral, etc$\ldots$ ). But how can I compute my integral $?$.
Let $$ f(a) = \int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x, $$
First, we note that $f(a)=f(-a)$, hence $f$ is even and we will focus only on $a \ge 0$.
Second, $$ \begin{split} \left| f(a) \right| &\le \left|\int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right|\\ &\le \int_0^{+\infty} \frac{\left|\cos(ax)\right|}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \\ &\le \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}} = f(0) \end{split}$$
With some math one finds that $$ f(0) = \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}} =2\int_0^{+\infty} \frac{\mathrm dt}{1+t^4} = \frac{\pi}{\sqrt{2}} \approx 2.22 $$
(see here)
Now, if we differentiate twitce under integral symbol (we can do it because of Leibniz integral rule) $$ \begin{split} f''(a) &= -\int_0^{+\infty} \frac{\cos(ax) \cdot x^2}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\ &= -\int_0^{+\infty} \frac{\cos(ax) \left(x^2+1-1\right)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\ &= - \left[\int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x -\int_0^{+\infty} \frac{\cos(ax) }{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right] \\ &= f(a) - \int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x \end{split}$$
Now,
$$ \begin{split} \int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x &= \frac{1}{2}\int_\mathbb{R}\frac{\cos(ax)}{\sqrt{|x|}} \, \mathrm{d} x \\ &= \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) \\ &=\frac{1}{2} \sqrt{\frac{2\pi}{|a|}}\\ &= \sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}} \end{split} $$ where we used $ \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) = \frac{1}{2} \sqrt{\frac{2\pi}{|a|}}$ (see Fourier Transform of $\frac{1}{\sqrt{|x|}}$)
So, the problem is now to solve
$$ \begin{cases} f''(a) -f(a) = -\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}}\\ f(0) = \frac{\pi}{\sqrt 2}\\ \end{cases} $$
From linear differential equations we know that a solution is
$$ f(a) = c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + f_\mathrm p(a) $$
where $c_1$ and $c_2$ are real numbers and $f_\mathrm p$ is the complementary solution.
Wolfram Alpha helps us to know that
$$ \begin{split} f(a) &= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\mathrm{e}^{-x+a}-\mathrm{e}^{x-a}}{2\sqrt{x}}\, \mathrm d x\\ &= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x \end{split} $$
Now, if we add the initial condition, we get $$ c_1+c_2 = \frac{\pi}{\sqrt 2}, $$ hence, after renaming the constant $c_1 = C$, the solution is
$$ \begin{split} f(a)&= C \mathrm{e}^a + \left(\frac{\pi}{\sqrt 2} - C\right) \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x\\ &= 2C \sinh(a) + \frac{\pi}{\sqrt 2} \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x \end{split} $$
If you find some other initial value (for example $f(1)$), then you can eliminate the constant $C$ too.
