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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!

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    $\begingroup$ $ \frac{1}{(49+x) \sqrt{x}}$ is undefined on $(-\infty, 0)$ $\endgroup$ Jul 31, 2017 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? $\endgroup$
    – user307169
    Jul 31, 2017 at 23:49
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    $\begingroup$ What is it with people in 2017 forgetting that complex numbers exist. $\endgroup$
    – Peyton
    Jul 31, 2017 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 $\endgroup$ Jul 31, 2017 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. $\endgroup$
    – user307169
    Jul 31, 2017 at 23:57

2 Answers 2

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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

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  • $\begingroup$ Integrate[1/(49 + x)/Sqrt[x], {x, -Infinity, 0}, PrincipalValue -> True] in Mathematica yields $0$. $\endgroup$
    – robjohn
    Aug 1, 2017 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, 2017 at 12:16
  • $\begingroup$ I agree zero is the answer. $\endgroup$
    – GEdgar
    Aug 1, 2017 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, 2017 at 12:59
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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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