# 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
5 years, 11 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. :)

- 5 years, 10 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!!! $\ddot\smile$

- 5 years, 10 months ago

Thanks Karthik! :D

I am going to post more of these. Meanwhile, you can try this one too: Trigonometric Integral. :)

- 5 years, 10 months ago

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.

- 5 years, 10 months ago

$I(p)= \int_0^1 \frac{x^{p-1} \rm li(x)}{\sqrt{-\ln x}} \ \mathrm{d}x\overset{x=e^{-u^2}} = 2\int_0^{\infty} e^{-pu^2} \rm li(e^{- u^2})\mathrm{d}u=\frac{2}{\sqrt{p}} \int_0^{\infty} e^{-u^2} \rm li(e^{-\frac{u^2}{p}}) \ \mathrm{d}u$ 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 $\int_0^{\infty} u^{2n} e^{u^2} \ \mathrm{d}u =\frac{(2n+1)!!}{2^n}$

- 5 years, 11 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.

- 5 years, 11 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.

- 5 years, 11 months ago

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. :)

- 5 years, 11 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.

- 5 years, 11 months ago

@Haroun Meghaichi : I have posted the solution, you may check. :)

- 5 years, 10 months ago

Fun Fact: This can also be written as $-2\sqrt{\frac\pi p}\operatorname{arcsinh}(\sqrt p)$.

- 5 years, 11 months ago