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:


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!

  • 4
    $\begingroup$ $ \frac{1}{(49+x) \sqrt{x}}$ is undefined on $(-\infty, 0)$ $\endgroup$ – steven gregory Jul 31 '17 at 23:48
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    $\begingroup$ The issue you describe actually occurs before that. The originally given integral doesn't even make sense. How can the expression $\dfrac1{(49+x)\sqrt x}$ be integrated over $(-\infty, 0)$ when $x \le 0$ is not allowed at all? Is it possible the integral you were given contains a typo? $\endgroup$ – tilper Jul 31 '17 at 23:49
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    $\begingroup$ What is it with people in 2017 forgetting that complex numbers exist. $\endgroup$ – Peyton Jul 31 '17 at 23:53
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    $\begingroup$ @Peyton Haha, I thought of that as I wrote it, however I knew taking it into the complex numbers was beyond the context in which the problem was given $\endgroup$ – amazonprime Jul 31 '17 at 23:56
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    $\begingroup$ @Peyton, this question doesn't have any tags that suggest complex numbers are involved. And complex numbers aren't part of the standard calculus curriculum. $\endgroup$ – tilper Jul 31 '17 at 23:57

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

  • $\begingroup$ Integrate[1/(49 + x)/Sqrt[x], {x, -Infinity, 0}, PrincipalValue -> True] in Mathematica yields $0$. $\endgroup$ – robjohn Aug 1 '17 at 3:03
  • $\begingroup$ int(1/(49+x)/sqrt(x), x = -infinity .. 0, CauchyPrincipalValue=true); in Maple yields $-\pi/7$. Strange how an integrand with real part zero everywhere could have principal part integral with nonzero real part... $\endgroup$ – GEdgar Aug 1 '17 at 12:16
  • $\begingroup$ I agree zero is the answer. $\endgroup$ – GEdgar Aug 1 '17 at 12:31
  • $\begingroup$ @GEdgar: I hate to disagree with Maple. I agree with Mathematica mainly because I computed the integral above. $\endgroup$ – robjohn Aug 1 '17 at 12:59

I swapped the extreme of integration so I got $$-\int_0^t \frac{1}{(x+49) \sqrt{x}} \, dx=-\frac{2}{7} \lim_{t\to \infty } \, \tan ^{-1}\left(\frac{\sqrt{t}}{7}\right)=-\frac{\pi }{7}$$


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