calculus - Value of $int_{- infty}^0 frac{mathrm dx}{(49+x) sqrt{x}}$?

I've been trying to find this integral:

$$\int_{- \infty}^0 \frac{\mathrm dx}{(49+x) \sqrt{x}}$$

I used a trigonometric substitution, and after some work arrived at this:

$$\arctan\left(\frac{\sqrt{x}}{7}\right)$$

So if written properly, it would look like:

$$\lim_{k \to -\infty} \arctan\left.\left(\frac{\sqrt{x}}{7}\right) \right|^{\,0}_{\,k}$$

Evaluating zero wasn't an issue, however taking the limit was. How can k approach negative infinity if it'll appear in a square root?

Wolfram Alpha told me this integral did not converge.
Another online calculator told me it was zero.
The person who gave this problem said it converged.

Any ideas? Thank you for your input!

Answer

This integral does not converge as either a Riemann or Lebesgue integral.

However, the integral does exist in the Cauchy Principal Value sense:

$$\newcommand{\PV}{\mathrm{PV}} \begin{align} \PV\int_{-\infty}^0\frac{\mathrm{d}x}{(49+x)\sqrt{x}} &=-2i\,\PV\int_0^\infty\frac{\mathrm{d}x}{49-x^2}\tag{1}\\ &=\frac{i}7\,\PV\int_0^\infty\left(\frac1{x-7}-\frac1{x+7}\right)\,\mathrm{d}x\tag{2}\\ &=\lim_{L\to\infty}\frac{i}7\,\PV\int_0^L\left(\frac1{x-7}-\frac1{x+7}\right)\,\mathrm{d}x\tag{3}\\ &=\lim_{L\to\infty}\left(\frac{i}7\,\PV\int_{-7}^{L-7}\frac1x\,\mathrm{d}x\right)-\lim_{L\to\infty}\left(\frac{i}7\int_7^{L+7}\frac1x\,\mathrm{d}x\right)\tag{4}\\ &=\lim_{L\to\infty}\left(\frac{i}7\,\PV\int_{-7}^7\frac1x\,\mathrm{d}x\right)-\lim_{L\to\infty}\left(\frac{i}7\int_{L-7}^{L+7}\frac1x\,\mathrm{d}x\right)\tag{5}\\[9pt] &=0\tag{6} \end{align}$$
Explanation:
$(1)$: substitute $x\mapsto-x^2$
$(2)$: partial fractions
$(3)$: write improper integral as a limit
$(4)$: substitute $x\mapsto x+7$ on the left and $x\mapsto x-7$ on the right
$(5)$: subtract the integral on $[7,L-7]$ from both limits
$(6)$: evaluate integrals