Who's up to the challenge? 55

Calculus Level 5

$\large \int _{ 0 }^{ \infty }{ \dfrac { { x }^{ 2 }\ln { (2x) } }{ { e }^{ x } } \, dx } =\ln { a } +b-c\gamma$

If the above equation holds true for positive integers $a,b$ and $c$ , where $\gamma$ denotes the Euler-Mascheroni constant , find $a+b+c$ .

The answer is 9.

3 solutions

Mark Hennings
Jun 7, 2017

The integral is $\Gamma(3)\ln 2 + \Gamma'(3) \; = \; \Gamma(3)\big[\ln2 + \psi(3)\big] \; = \; 2\ln2 + 3 - 2\gamma \; = \; \ln4 + 3 - 2\gamma$ making the answer $4+3+2=\boxed{9}$ .

First Last
May 12, 2016

Log rule! $\displaystyle\ln 2 \times \int_{0}^{\infty}\frac{x^2}{e^x}dx + \int_{0}^{\infty}\frac{x^2\ln(x)}{e^x}dx$

$\displaystyle\ln 2 \times \int_{0}^{\infty}\frac{x^2}{e^x}dx = \ln(4)$

The second integral is more interesting:

$\displaystyle\int_{0}^{\infty}\frac{x^2\ln(x)}{e^x}dx = \frac{-x^2-2x-2}{e^x}\ln x |_0^{\infty} + \int_{0}^{\infty}\frac{x+2}{e^x}dx + \frac{2}{x e^x}dx =$ , by integration by parts.

$\displaystyle 2 \lim_{x\to 0}( \ln x - Ei(x)) + 3 = -2\gamma + 3$

$\displaystyle\int_{0}^{\infty}\frac{x^2\ln(2x)}{e^x}dx = \boxed{\ln{4} + 3 - 2\gamma}$

Help me out!!! $let\quad I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ 2 } }{ { e }^{ x } } } \log { ax } dx\\ \frac { dI }{ da } =\int _{ 0 }^{ \infty }{ \frac { { x }^{ 2 } }{ { e }^{ x } } } \times \frac { a }{ ax } dx\\ \frac { dI }{ da } =\int _{ 0 }^{ \infty }{ \frac { x }{ { e }^{ x } } } dx\\ \frac { dI }{ da } =\quad 1\\ \quad dI\quad =\quad da\\ I\quad =\quad a\quad +\quad c\\ I\quad =\quad 2\quad +\quad c$ what shall I do after this?? please help!!

@Calvin Lin and others..please see to it

Righved K - 5 years, 1 month ago

What you've done is define $I$ as a constant which it is not. When you differentiate $I$ with respect to $a$ , then integrate later, $I$ is still a function of $a$ .

Your final 2 statements should be: $\displaystyle I(a) = a + c,\quad I(2) = 2 + c$

Because there is no easy $a$ value to plug in originally to get a constant, there is no use for Leibniz rule here. Plugging in anything greater than zero for a would result in the same process to find $c$ as it would to actually evaluate the integral in the first place. Here is a good video I found explaining Leibniz rule: https://goo.gl/YfwbFP

First Last - 5 years, 1 month ago

I(the integral) would definitely be a constant value(since it is a definite integral),Moreover the statements that I(me :P) have written are valid..I got through the video(you are right, the constant is difficult to be found) but I as I have approached,there can be a way to solve it, isn't it?..please others and especially @Calvin Lin ..please see to it:)

Righved K - 5 years, 1 month ago

@Righved K – Like I said, you need to find an $a$ to plug into $I(a)$ originally to get a simple result to later computer $c$ . This can't be done in this case so finding this $a$ would be equivalent to solving the specific integral. So there is no use for Leibniz rule.

When I said constant, I meant your "function", $I$ , which should be $I(a)$ .

First Last - 5 years, 1 month ago

@First Last – Got you. Thank-you so much!:D

Righved K - 5 years, 1 month ago

As pointed out by Jasper, we're looking at the function $I(a)$ . In particular, we are differentiating with respect to $a$ .
In your second line, what is the value of $\frac{ d \log ax} { da}$ ? Hint: It is not $\frac{a}{ax}$ .
After you fix that step, as pointed out by Jasper, you need to determine the value of $c$ . This is where most of the work is involved, as seem in the "the second integral is more interesting" comment of the solution. I currently do not see an easy way for Righved's solution to proceed without explicitly performing the integral (but I could be wrong).

Calvin Lin Staff - 5 years, 1 month ago

Thanks @Calvin Lin, yes I am sorry(I made a school boy error) , it would be 1/a and yes you are right, even I too couldn't find any easy path to proceed in my solution. I could only work out for the ln4 part of the solution :(

Righved K - 5 years, 1 month ago

@Righved K – Right, so we arrived at $I(a) = 2\ln a + I(1)$ , which could be obtained directly since $\int \frac{ x^2}{e^x} = 2$ .

Having said that, don't be discouraged. Differentiating through the integral is a useful approach where we already know the value at a given point and want to extend out along the family of curves.

Calvin Lin Staff - 5 years, 1 month ago

@Calvin Lin – @Calvin Lin ..thankyou so much and please edit your reply ( the value of integral would be 2) cheers!!!....thank you so much once again!!!

Righved K - 5 years, 1 month ago

Guilherme Niedu
Jun 1, 2017

$I = \large \displaystyle \int_0^{\infty} x^2 \ln(2x) e^{-x} dx$

Integrating by parts:

$\large \displaystyle I = - \left [ e^{-x} x^2 \ln(2x) \right ] \Bigg |_0^{\infty} + \int_0^{\infty} \left [ x + 2x \ln(2x) \right ] e^{-x} dx$

$\large \displaystyle I = \int_0^{\infty} \left [ x + 2x \ln(2x) \right ] e^{-x} dx$

Integrating by parts again:

$\large \displaystyle I = - \left [ e^{-x} ( x+ 2x\ln(2x) \right ] \Bigg |_0^{\infty} + \int_0^{\infty} \left [ 1 + 2 \ln(2x) + 2 \right ] e^{-x} dx$

$\large \displaystyle I = \int_0^{\infty} 3 e^{-x} dx + 2 \int_0^{\infty} e^{-x} \ln(2x) dx$

$\large \displaystyle I = 3 + 2 \left[\int_0^{\infty} e^{-x} \ln(2) + \int_0^{\infty} e^{-x} \ln(x)dx \right ]$

$\large \displaystyle I = 3 + 2\ln(2) + 2\int_0^{\infty} e^{-x} \ln(x)dx$

Since $\int_0^{\infty} e^{-x} \ln(x)dx = -\gamma$ :

$\color{#20A900} \boxed{\large \displaystyle I = \ln(4) + 3 -2\gamma }$

Thus:

$\color{#3D99F6} \large \displaystyle a=4, b=3, c=2, \boxed{\large \displaystyle a+b+c=9}$