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Why are these integrals the same?

$$ \int_0^1\bigg(\frac{-1}{\ln^3(x)}\bigg)\exp\bigg(\frac{1}{\ln(x)}\bigg)~dx=\int_0^1 \bigg(-\ln(x)\bigg)\exp\bigg(\frac{1}{\ln(x)}\bigg)~dx=2K_2(2) $$

Where the $K_2$ is a modified Bessel function.

Is there some property that describes when this happens?

I think it has to do with the functions being in the same general form, but I can't pinpoint the specifics.

Thanks.

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  • $\begingroup$ This is quite nice and it's not hard to show that they are equal (by easy, I mean elementary without using some special functions). How did you find those integrals? $\endgroup$ – Zacky Oct 31 '19 at 23:19
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    $\begingroup$ @Nyssa I knew that $\int_0^1 \exp(\frac{1}{\ln(x)})dx=2K_1(2)$ and I wanted to find similar integrals for $2K_n(2)$ $n=0,2,3,...$ Then I noticed that pairs of integrals had the same area. $\endgroup$ – geocalc33 Oct 31 '19 at 23:27
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We can show that the difference between them is $0$ which shows how they are equal. $$\int_0^1 \exp\left(\frac{1}{\ln x}\right)\left(\ln x-\frac{1}{\ln^3 x}\right)dx\overset{\ln x \to x}=\int_{-\infty}^0 e^{x+1/x}\left(x-\frac{1}{x^3}\right)dx$$ $$=\int_{-\infty}^0 e^{x+1/x}\left(1-\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)dx=\int_{-\infty}^0 \left(e^{x+1/x}\right)'\left(x+\frac{1}{x}\right)dx$$ $$\overset{IBP}=-\int_{-\infty}^0 e^{x+1/x}\left(1-\frac{1}{x^2}\right)dx=-e^{x+1/x}\bigg|_{-\infty}^0=0$$


And if I'm not making a mistake this can be generalized to: $$\int_0^1 \exp\left(\frac{1}{\ln x}\right)\ln^{n-1} xdx=\int_0^1 \exp\left(\frac{1}{\ln x}\right)\frac{1}{\ln^{n+1}x} dx\rightarrow 2(-1)^{n+1}K_n(2)$$

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    $\begingroup$ Thanks this is helpful $\endgroup$ – geocalc33 Oct 31 '19 at 23:43

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