Prove the following:

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

Prove the following:

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

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TopNewestFirst, 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 · 2 years, 3 months ago

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– Karthik Kannan · 2 years, 3 months ago

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!!! \(\ddot\smile\)Log in to reply

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

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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 · 2 years, 3 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 · 2 years, 3 months ago

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– Pranav Arora · 2 years, 3 months ago

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.Log in to reply

– Haroun Meghaichi · 2 years, 3 months ago

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.Log in to reply

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 · 2 years, 3 months ago

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– Ali Caglayan · 2 years, 3 months ago

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.Log in to reply

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

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