Integral of ln(x)*tan^-1(x) from 0 to 1

Integral of ln(x)*arctan(x) from 0 to 1,

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[GreenMeansGOF]

Awesome. This was my integral. I did it differently and now I see that your way is easier. Awesome video as always.

[WhiteDotX]

glad to see my comment in the intro! After 3 years, now you're near to a million subs, congrats!

[thenateman]

Summation and integration are interchangeable because both are linear (the integral of a sum of functions is the same as a sum of the integrals of those functions), and because both functions are nice enough in the interval (no discontinuities, not piecewise, etc...). :)

[emperorpingusmathchannel]

I saw you stuttering but don't worry about it.






It was actually not that bad.

[toasticide]

Can I just say how much I love your channel. Before I started my degree your videos helped refine my basic knowledge on integrals. And now even after I learn about expansions I learn of new ways to solve integrals!

[VaradMahashabde]

The answer may have three terms in it, but they are all constants, so it's actually not that bad. What is also not bad is that your channel's videos are like a yet-to-be-compiled calculus course book

[stydras]

Because the partial sums of tan⁻¹'s powerseries converges to a continous function on [-1, 1], every part of the summation multiplied with ln|x| is riemann integrable. Therefore we can switch integation and (infinite) addition.

Also ln|x| turns into ln(x) because we only care about [0, 1]

:)

[Roarshark]

Wow what a satisfying integral! I love how you incorporated summation and limits in deriving the solution (as well as integration by parts).

[GreenMeansGOF]

Me: How are you doing today?

bprp: Check the video description to watch another one of my videos where I discuss how I’m doing.

[domanicmarcus]

I love how you edited your video. It all goes downhill from 12:03 until 12:26. By the way, you TRULY are the GREATEST MATH TEACHER that I have EVER SEEN. May God ALWAYS BLESS YOU.

[max-yasgur]

I think he broke. Someone change his batteries.

[sugarfrosted]

16:17 You could also see why it's true by looking at the power series for inverse tangent given on the other side of the board. :p

[ralfbodemann]

For an easier approach, just watch my video in the description below. It's actually not that bad.
SeriEsly: Thanks, Steve! Great job again!

[jonathangrey]

You can take the summation out because it doesn’t have an x term, meaning it can come out and it won’t affect integration.

[thesinglemathnerd]

I randomly came up with a different but somewhat alike integral in my head after seeing this. It is the integral from 0 to pi/4 of ln(tan(x))*tan(x) which happens to be -pi^2/48. The way I did it involves a total of like three of substitutions, THREE applications of L’Hopital’s rule (yes, three on the same limit), and some use of series for the ln function.

[wiwaxiasilver]

2 years later, *it’s actually not that bad*
But like seriously, loved it as usual.

[kaksihapsi]

I believe we can save some time by integrating by parts right at the beginning (we need to integrate the term tan^{-1}(x) (again by parts) and derive the term ln(x)). We get that the result -pi/4 + 1/2 * ln(2) + 1/2 * int_0^1 ln(x^2+1)/x dx, which is just -pi/4 + 1/2 ln(2) -1/4 * Li_2(-1). The value of the last integral can be calculated by summing a series (just as in the video).

[AdityakrishnaMr]

12:20 Well that's actually not that bad!

[Bignic]

On the interval [0, 1], the function ln(x)*(-1)^n*x^(2n+1)/(2n+2) is dominated by e^x, because e^0=1 and is increasing forever, while the function in the sum starts at 0. So you can interchange the infinite sum and integral with the dominated convergence theorem.

[dreadk]

There are different type of conditions for interchanging the integral with the sum. One states that if f_n is a sequence of continuous functions on an interval [a, b] and the functional series of f_n(x) from n=1 to infinity uniformly convergences then we can interchange the integral with the sum. In this case we do have continuous functions but the uniform convergence needs to be proved separately.