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An interesting double definite integral

Prove the following:

\[\int_0^1 \int_0^x \frac{1}{\ln t}\frac{x^{p-1}}{\sqrt{\ln\frac{1}{x}}}\,dt\,dx=-2\sqrt{\frac{\pi}{p}}\ln\left(\sqrt{p}+\sqrt{p+1}\right) \]

Note by Pranav Arora
3 years, 2 months ago

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First, to make it simpler to write, define:

\(\displaystyle \text{li}(x)=\int_0^x \frac{1}{\ln t}\,dt \)

\(\text{li}(x)\) (Logarithmic Integral ) is termed as a special function but the following solution will not use any special properties related to this function.


\[\displaystyle \int_0^1 \text{li}(x) \frac{x^{p-1}}{\sqrt{\ln\frac{1}{x}}}\,dx \]

Use the substitution \(x=e^{-u^2} \Rightarrow dx=-2ue^{-u^2}\,du\).

\(\displaystyle \Rightarrow I(p)=2\int_0^{\infty} \text{li}(e^{-u^2})e^{-pu^2}\,du\)

Divide both the sides by two and use integration by parts to obtain:

\(\displaystyle \frac{I(p)}{2}=\left(u\text{li}(e^{-u^2})e^{-pu^2}\right|_0^{\infty}-\int_0^{\infty} u\left(\text{li}(e^{-u^2})(-2pue^{-pu^2})+\frac{e^{-pu^2}(-2ue^{-u^2})}{\ln(e^{-u^2})}\right)\,du\)

\(\displaystyle \Rightarrow \frac{I(p)}{4}=p\int_0^{\infty} u^2\text{li}(e^{-u^2})e^{-pu^2}\,du-\int_0^{\infty} e^{-(p+1)u^2}\,du\)

\(\displaystyle \Rightarrow \frac{I(p)}{4}=-p\frac{I'(p)}{2}-\frac{\sqrt{\pi}}{2\sqrt{p+1}}\)

\(\displaystyle \Rightarrow I'(p)+\frac{I(p)}{2p}=-\frac{\sqrt{\pi}}{2p\sqrt{p+1}}\)

The above is a linear differential equation and its solution is:

\(\displaystyle I(p)=-2\sqrt{\frac{\pi}{p}}\ln\left(\sqrt{p}+\sqrt{p+1}\right)+C\)

As \(p\) tends to infinity, \(I(p)\) tends to zero, hence \(C=0\).

\[\displaystyle \Rightarrow I(p)=-2\sqrt{\frac{\pi}{p}}\ln\left(\sqrt{p}+\sqrt{p+1}\right)\]


Please feel free to ask me any doubts regarding the solution. :) Pranav Arora · 3 years, 2 months ago

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@Pranav Arora Great solution Pranav! This is indeed an extremely difficult problem (at least for me). I tried many things but was unsuccessful. Each and every step is just awesome!!! \(\ddot\smile\) Karthik Kannan · 3 years, 2 months ago

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@Karthik Kannan Thanks Karthik! :D

I am going to post more of these. Meanwhile, you can try this one too: Trigonometric Integral. :) Pranav Arora · 3 years, 2 months ago

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@Karthik Kannan Great solution. the last limit is \(0\) by Lebesgue DCT because the integrand is unbounded (therefore not Riemann integrable) but it still Lebesgue integrable.

If you don't want anything fancy, you can just plug \(p=1\) and calculate the integral by parts to arrive to \(-2\sqrt{\pi}\) and the rest follows. Haroun Meghaichi · 3 years, 2 months ago

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\[I(p)= \int_0^1 \frac{x^{p-1} \rm li(x)}{\sqrt{-\ln x}} \ \mathrm{d}x\overset{x=e^{-u^2}} = 2\int_0^{\infty} e^{-pu^2} \rm li(e^{- u^2})\mathrm{d}u=\frac{2}{\sqrt{p}} \int_0^{\infty} e^{-u^2} \rm li(e^{-\frac{u^2}{p}}) \ \mathrm{d}u\] Use the series described in the logarithmic integral wiki page to get : \[I(p) = \frac{2}{\sqrt{p}} \int_0^{\infty} e^{- u^2} \left(\gamma + \ln \frac{u^2}{p} +\sum_{n\geq 1 } \frac{(-1)^n p^{-n}u^{2n}}{n\cdot n!}\right) \ \mathrm{d}u\] Now, it is just expand and Taylor series to finish (too lazy to finish it) but here's the main idea : \[\int_0^{\infty} e^{-u^2}(\gamma +\ln \frac{u^2}{p} ) \mathrm{d}u = -\frac{-\sqrt{\pi}\ln(4p)}{2} \] and \[\int_0^{\infty} u^{2n} e^{u^2} \ \mathrm{d}u =\frac{(2n+1)!!}{2^n} \] Haroun Meghaichi · 3 years, 2 months ago

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@Haroun Meghaichi Nice approach but for the current problem, you don't really need to use such heavy machinery. There exists a very very elementary solution. Its been a week already but I think I should wait some more. Pranav Arora · 3 years, 2 months ago

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@Pranav Arora I really want to see an elementary approach (I guess that it starts with integration by parts). BTW, why there is not much level 4 and 5 problems in Calculus ? I have just posted a limit and an easy integral, I hope that you like them. Haroun Meghaichi · 3 years, 2 months ago

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@Haroun Meghaichi Yes, integration by parts is the way to go but there's more to it.

I myself do not understand why there are not many problems in level 4 or 5. Posting the integrals as a problem doesn't help much because most of times, W|A or Mathematica are able to evaluate them. This is the reason I have posted the problems as notes.

The problems you posted are nice, I think I am onto something for the integral. I will post a complete solution tomorrow. Too tired at the moment. :) Pranav Arora · 3 years, 2 months ago

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@Pranav Arora Rather than getting a straight answer for the integral add a layer of problem over it so that it cannot be blindly typed into WA. Ali Caglayan · 3 years, 2 months ago

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@Haroun Meghaichi @Haroun Meghaichi : I have posted the solution, you may check. :) Pranav Arora · 3 years, 2 months ago

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Fun Fact: This can also be written as \(-2\sqrt{\frac\pi p}\operatorname{arcsinh}(\sqrt p)\). Ali Caglayan · 3 years, 2 months ago

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