Challenge 31: Can You Evaluate This Crazy Integral?!!

Congratulations to Aryan Soman, Plasmaboy, Dingchen Wei, Kyro, TheRedhotbr, The Kirtpole, Sighmaniac RotMG, Brain it on!, Jupiterninja95, and laxman zanwar for successfully solving the last week's math challenge question! Aryan Soman was the first person to solve the question.

Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

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[rohannuckchady]

substitute u = x/2018 and the integral evaluates to (2018)^(a-1)/(a-1)!, a= alpha. then change the sun from 1 to infinity to 0 to infinity, so u get a sum of (2018)^n/n! which is e^2018, ln of that gives 2018

[blurryfca]

thanks for such a good question.... wonderful n working on the beautifully laid out powers of x and the exponential function helping all along the way....the end part was the mostt fun when i got was the taylor series of e^x (x=2018) absolutely loved it....keep posting such beautiful sums😊

[fakeaccount]

By factoring out some constants and using a u substitution, rewriting the integral will give the gamma function. The sum of the integrals reduces to the Taylor series expansion of e^2018, giving a final answer of 2018.

[sunijshah]

First evaluate the integral x^(a-1)*e^(-x/2018)...by using integration by parts...that is differentiate x^(a-1) and integrate the second part...after applying it many times we get a series and the integral simplifies to now we get summation of [2018^(a-1)]/(a-1)!....this is the series expansion of e^x with x = we get the answer....2018

[coolclips]

Beautiful problem. I'll spend some time thinking about this, but I'm pretty sure I can solve it.

[jacobryan]

I let u = x/2018 and pulled constants out of the integral. The noticed the integral is now gamma(alpha). Use relationship gamma(alpha) = factorial(alpha-1) and cancel with denominator. Shift sum to start at alpha=0 by adding 1 to alpha in the expression. Notice this is a Taylor series for e^x with x=2018 and simplify with the natural log to give an answer of 2018.

[dorianvoydie]

We substitute : u=x/2018 then we integrate by parts (k-1) times, so we have (k-1)! * int(e^-u)du from 0 to inf, which equals to 1. The factorial on numerator cancels with the square in denominator, we recognise a Taylor serie expansion : e^2018 ==> composing by ln equals to [2018]

[smitashripad]

Doing the substitution of x/2018 and expanding a bit using gamma function we get e^2018 which after ln becomes 2018 (also using the definition of e^x ie Taylor series).

[curiousminds]

First we will let u=x/2018 and then du = dx/ 2018 which is there then we will take out constants with alpha and what we will left in the integral is gamma function now simplifying the sum using tyalor series gives e^2018 and ln(e^2018) = 2018

[abdulazizalbaiz]

I am a huge fan of these weekly challenges. So sad that I just found about them recently. But I must say, putting answers in YouTube comment section is not the best way to gather answers, as it reveals the correct answer the moment it is posted.

Suggestion: create an email address solely for this purpose.

[FedericoYulita]

answer is 2018. The integral is (2018^alpha)*(alpha-1)! and so then the sum is just e^2018 because its the Taylor form of e^x and so ln(e^2018)=2018

[billmbogho]

First one ive been able to solve ... integral part smoothens into gamma function then we get series expansion of e^2018.... natural log is therefore 2018

[aaronlamoreaux]

The answer is 2018. I did u-substitution with u = x / 2018 and then integration by parts in terms of alpha in order to get the integral as 2018 ^ (alpha - 1) / (alpha - 1)!. The infinite sum is just a power series for e^x which natural log gives 2018 as the answer.

[nishantreddy]

On substituting x/2018 = t, we have a gamma function which evaluates to (a-1)! [Here a is aplha] . Then we get the expansion of e^2018. Hence the ans is 2018.

[QuoteVG]

Log[Sum[Integrate[x^(a-1)*E^(-x/2018)/(2018((a-1)!)^2), {x, 0, Infinity}], {a, Infinity}]]

= 2018.
After all of this hard work, we have finally arrived at the solution. We are done.

[Nabil_Bennai]

At first get the denominator outside the integral out, so you get the integral of x^(a-1)exp(-x/2018) from 0 to Infinity = -2018^a Gamma(a) so we have
ln
(a-1)!=Gamma(a)
ln SUM(2018^(a-1)/Gamma(a))
ln exp(2018)
2018

[picleus]

I got 2018 using a slow method (haven't learned about the gamma function or anything of that sort yet). First I factored out the denominator of the integral, as it is constant on any α. Then I used tabular method of integration by parts to create a sum for the integral. Did a bit of manipulation and ended with the integral equaling to [(2018)^(α-1)]/[(α-1)!]. The infinite sum of this gives e^(2018), and ln(e^(2018)) is just 2018. (I also wrote everything in pen for an extra-hardcore math challenge.) ;-)

[IustinThe_Human]

After letting x= 2018 u, changing some thing around you get the intregal get the form of the definition of factorial so you have in the end ln of infinte sum from (α=1) of 2018^(α-1)/(α-1)! which is ln(exp(2018) ) and it is of course 2018

[colorfulcalculus]

First, let’s have a u-sub with u = x/2018. Next, after u-sub, we will see that after moving independent variables outside of the integral, we will have a gamma function.
Next, the infinite series is one of the Taylor expansion for e^(2018x). Thus, the answer is ln[e^2018] = 2018

[nickworms]

Well I didn't use the gamma function, and just found the first four terms unsimplified, and noticed a 2018 increasing in power with the term from the u substitution. i also noticed a factorial multiplying from the tabular integration that i used. Then there was the factorial squared on the bottom which would cancel with the one on top leaving just a single factorial. I then simplified the original series to my new series made from that pattern which seemed similar to the Maclaurin series for e^x but with 2018 substituted for x. Thus the inside part is the same as e^2018. and ln(e^2018) is 2018. Larger and harder than other people but I never really learned about the gamma function in my AP BC Calc course this year. R I P