Let's do some calculus! (49)

Calculus Level 5

$\begin{aligned} & \large I_1 = \int_0^{\infty} \dfrac{\ln^2 (x)}{e^x} x^{{1}/{4}} \mathrm{d}x \\ & \large I_2 = \int_0^{\infty} \dfrac{x^{-{1}/{4}}}{e^x} \mathrm{d}x \end{aligned}$

Let $I_1$ and $I_2$ be as defined above. If the closed form of $I_1 I_2$ can be represented as

$\dfrac{a}{b \sqrt c} \pi^3 - \sqrt{c} \pi^2 + \dfrac{d}{c \sqrt{c}} \pi^2 \ln(c) + \dfrac{\pi^2 \gamma}{c \sqrt c} + \dfrac{\pi \gamma^2}{c \sqrt c} + \gamma \pi \left( \dfrac{d \ln(c)}{\sqrt c} - c \sqrt c \right) - e \sqrt{c} \pi \ln(c) + \dfrac{f \pi \ln^2 (c)}{c \sqrt{c}} + c \sqrt{c} \pi G$

where alphabets in lowercase from $a$ to $f$ are distinct positive integers with coprime-pairs $(a,b) , (c,d)$ and $(c,f)$ and $c$ is not a perfect power, evaluate $a+b+c+d+e+f$ .

Notations:

$G$ denotes the Catalan's constant .
$\gamma$ denotes the Euler-Mascheroni constant .
$\ln^n (\cdot) = {(\ln (\cdot))}^n$

Inspiration.

For more problems on calculus, click here .

The answer is 33.

1 solution

Mark Hennings
Jun 10, 2017

We note that $I_1 \; = \; \Gamma''\big(\tfrac54\big) \; = \; \Gamma\big(\tfrac54\big)\big[\psi\big(\tfrac54\big)^2 + \psi'\big(\tfrac54\big)\big] \hspace{3cm} I_2 \; = \; \Gamma\big(\tfrac34\big)$ and so $\begin{aligned} I_1I_2 & = \; \Gamma\big(\tfrac54\big)\Gamma\big(\tfrac34\big)\big[\psi\big(\tfrac54\big)^2 + \psi'\big(\tfrac54\big)\big] \; = \; \tfrac14\Gamma\big(\tfrac14\big)\Gamma\big(\tfrac34\big)\big[\psi\big(\tfrac54\big)^2 + \psi'\big(\tfrac54\big)\big] \\ & = \; \tfrac14\pi \mathrm{cosec}\,\tfrac14\pi\big[\psi\big(\tfrac54\big)^2 + \psi'\big(\tfrac54\big)\big] \; = \; \tfrac{\pi}{2\sqrt{2}}\big[\psi\big(\tfrac54\big)^2 + \psi'\big(\tfrac54\big)\big] \end{aligned}$ Now $\begin{aligned} -\gamma - 2\ln2 \; = \; \psi\big(\tfrac12\big) & = \tfrac12\psi\big(\tfrac14\big) + \tfrac12\psi\big(\tfrac34\big) + \ln 2 \; = \; \psi\big(\tfrac14\big) + \tfrac12\pi \cot\tfrac14\pi + \ln2 \\ & = \psi\big(\tfrac14\big) + \tfrac12\pi + \ln2 \end{aligned}$ and so $\psi\big(\tfrac54\big) \; = \; \psi\big(\tfrac14\big) + 4 \; = \; 4 - \gamma - 3\ln2 - \tfrac12\pi$ On the other hand $\psi'\big(\tfrac54\big) \; = \; \psi'\big(\tfrac14\big) - 16 \; = \; \pi^2 + 8G - 16$ and hence $\begin{aligned} I_1I_2 & = \frac{\pi}{2\sqrt{2}}\left[\big(4 - \gamma - 3\ln2 - \tfrac12\pi\big)^2 + \pi^2 + 8G - 16 \right] \\ & = \frac{\pi }{2\sqrt{2}}\Big[ \gamma^2 + 9(\ln2)^2 + \tfrac54\pi^2 + 8G - 8\gamma - 24\ln2 - 4\pi + 6\gamma\ln2 + \gamma\pi + 3\pi\ln2 \Big] \\ & = \begin{array}{l} \frac{5\pi^3}{8\sqrt{2}} - \sqrt{2}\pi^2 + \frac{3}{2\sqrt{2}}\pi^2\ln2 + \frac{\pi^2\gamma}{2\sqrt{2}} + \frac{\pi \gamma^2}{2\sqrt{2}} \\ + \gamma\pi\Big(\frac{3\ln2}{\sqrt{2}} - 2\sqrt{2}\Big) - 6\sqrt{2}\pi\ln2 + \frac{9\pi (\ln2)^2}{2\sqrt{2}} + 2\sqrt{2}\pi G \end{array} \end{aligned}$ This makes $a=5$ , $b=8$ , $c=2$ , $d=3$ , $e=6$ , $f=9$ , and so the answer is $5+8+2+3+6+9 = \boxed{33}$ .

Fantastic!

Tapas Mazumdar - 4 years ago

nice problem!, although the answer was a bit tedious. it would have been better if you asked the sum of the constants in $\dfrac{\pi}{c\sqrt{c}} \left(-a^2+bG+d\pi^2+\left(a-e\gamma-f\ln(c)-\dfrac{\pi}{c}\right)^2\right)$

Aareyan Manzoor - 3 years, 12 months ago

I had thought about that but wasn't very sure that the expression could not be manipulated (i.e., making the expression in the square different) but I find that it would not have been the case.

Also, if one can find the above form then one can also do expansion. It's basically just there for more thrill as readers may get intimidated by such a large expression.

Tapas Mazumdar - 3 years, 12 months ago

Let's do some calculus! (49)

For more problems on calculus, click here .

The answer is 33.

1 solution

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