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I recently came across the Frullani integral and wanted to try it out by seeing if solving $\int_0^\infty{\mathrm{sinc}(x)}$ would give the familiar answer of $\pi/2$, but instead I got $-\pi/2$. I did the following:

$$ \begin{align} \int_0^\infty{\frac{\sin x}{x}}\ dx &= \int_0^\infty{\frac{e^{ix}-e^{-ix}}{2ix}}\ dx = \frac{1}{2i}\int_0^\infty{\frac{e^{ix}-e^{-ix}}{x}}\ dx \\ &= \frac{1}{2i}\int_0^\infty{\frac{e^{-(-i)x}-e^{-(i)x}}{x}}\ dx \end{align} $$

Then, taking $f(x)=e^{-x}$, $a=-i$ and $b=i$, the solution becomes

$$ \begin{align} \frac{1}{2i}\int_0^\infty{\frac{e^{-(-i)x}-e^{-(i)x}}{x}}\ dx &= \frac{1}{2i}(e^{-\infty}-e^{-0})\ln \frac{-i}{i} \\ &= \frac{1}{2i}(-1)\ln(-1) \\ &= -\frac{1}{2i} \cdot i\pi = -\frac{\pi}{2} \end{align} $$

I believe the error may be due to my use of $\ln(-1)=i\pi$ being invalid in this scenario, but I am not certain about this. For now, I have found two ways to "correct" the formula so that the answer gives $\pi/2$ instead of $-\pi/2$:

  • Since $-1=\frac{1}{-1}$, I can deduce that $\ln(-1)=\ln((-1)^{-1})=-\ln(-1)=-i\pi$, and the two negatives cancel out. However, I'm pretty sure there's something very wrong with this and I should not be allowed to do this.
  • If I instead do: $$ \begin{align} \frac{1}{2i}\int_0^\infty{\frac{e^{ix}-e^{-ix}}{x}}\ dx &= \frac{1}{2i}\int_0^\infty{\frac{-(e^{-ix}-e^{ix})}{x}}\ dx \\ &= -\frac{1}{2i}\int_0^\infty{\frac{e^{-ix}-e^{ix}}{x}}\ dx \end{align} $$ and then take $f(x)=e^{-x}$, $a=i$ and $b=-i$, there are again two negatives that cancel out and give $\pi/2$. This works, but it still does not answer my question of why my first attempt didn't.
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  • $\begingroup$ In the reference you stated, f(x) should be a function defined for all non-negative real numbers and a and b are real too. $\endgroup$
    – Lai
    Apr 18, 2023 at 0:13
  • $\begingroup$ @Lai you're right, I read through it multiple times and still missed it, no clue how. The fact that it's so close to being correct still makes me wonder though. I did some further checking around and found math.stackexchange.com/questions/1807410/… ; could the answers here apply to this case? $\endgroup$ Apr 18, 2023 at 5:06
  • $\begingroup$ Thank you for your reference which extends the Frullani’s Theorem for complex numbers. $\endgroup$
    – Lai
    Apr 18, 2023 at 7:40

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