Tutorial on some interesting techniques involved while solving some tricky integrals

Question

vishweshshrimali5 · Answer

Evaluate:
$$\large{\int\limits_{0}^{1}[\ln(x)\ln(1-x)dx]}$$

Here is my solution:

Sol:

Let :

$$\large{I = \int\limits_{0}^{1}[\ln(x)\ln(1-x)dx]}$$

Now, the function $\large{\ln{x}\ln{1-x}}$ is non continuous at x = 0 and x = 1.

So, I will write I as,

$$\large{I = \lim_{\cfrac{\epsilon \rightarrow 0^{+}}{\delta \rightarrow 0^{+}}} \int\limits_{\epsilon}^{1-\delta}[\ln(x)\ln(1-x)dx]}$$ -----(1)

Now, for x satisfying:

$\large{0 < \epsilon \le x \le 1 - \delta < 1}$ and some positive integer $\large{n}$;

$$\large{-\ln(1-x) = \sum_{k=1}^{\infty}(\cfrac{x^k}{k})}$$

$$\large{= \sum_{k=1}^{n}(\cfrac{x^k}{k}) + x^{n+1}\sum_{k=n+1}^{\infty}(\cfrac{x^{k-(n+1)}}{k})}$$ ----------(2)

Thus,

$$\large{|\ln(1-x) + \sum_{k=1}^{n}{\cfrac{x^k}{k}}|}$$

$$\large{= x^{n+1}\sum_{k=n+1}^{\infty}{\cfrac{x^{k-(n+1)}}{k}}}$$

$$\large{= x^{n+1} \sum_{k=0}^{\infty}{\cfrac{x^k}{k+n+1}}}$$

$$\large{\le \cfrac{x^{n+1}}{n+1} \sum_{k=0}^{\infty}(x^k)}$$

$$\large{= \cfrac{x^{n+1}}{(n+1)(1-x)}}$$ ----------(3)

Thus,

$$\large{|\int\limits_{\epsilon}^{1-\delta}\ln{x}[\ln{(1-x)} + \sum_{k=1}^{n}{\cfrac{x^k}{k}}]dx|}$$

$$\large{\le \cfrac{1}{n+1} \int\limits_{\epsilon}^{1-\delta} (-\ln{x})*\cfrac{x^{n+1}}{(1-x)} dx}$$ ----------------(4)

Now, for $\large{y \ge 1$;

$$\large{0 \le \ln{y} = \int\limits_{1}^{y}\cfrac{dt}{t} \le \int\limits_{1}^{y}dy = y - 1}$$ ---------- (5)

Putting $\large{y = \cfrac{1}{x}}$ in (5),

$$\large{0 \le -\ln{x} = \ln(\cfrac{1}{x}) \le \cfrac{1}{x} - 1 = \cfrac{1-x}{x}}$$ --------(6)

Using, (6) in (4);

$$\large{|\int\limits_{\epsilon}^{1-\delta}\ln{x}\ln{1-x}dx + \sum_{k=1}^{n}\cfrac{1}{k}\int\limits_{\epsilon}^{1-\delta}\ln{x}x^{k}dx|}$$

$$\large{\le \cfrac{1}{n+1} \int\limits_{\epsilon}^{1-\delta}x^n dx}$$

$$\large{< \cfrac{1}{n+1} \int\limits_{0}^{1}x^n dx}$$

$$\large{= \cfrac{1}{(n+1)^2}}$$

Now, letting, $\large{n \rightarrow \infty}$,

$$\large{\int\limits_{\epsilon}^{1-\delta}\ln{x}\ln{1-x}dx}$$

$$\large{= -\sum_{k=1}^{\infty} \cfrac{1}{k}\int\limits_{\epsilon}^{1-\delta}\ln{x}x^{k}dx}\}$$
------------(7)

Now,

$$\large{\cfrac{d}{dx}(\cfrac{x^{k+1}\ln{x}}{k+1} - \cfrac{x^{k+1}}{(k+1)^2}) = x^k*\ln{x}}$$ ---------- (8)

Thus,

By fundamental theorem of calculus,

$$\large{|\int\limits_{\epsilon}^{1-\delta}\ln{x}x^k dx}$$

$$\large{= [\cfrac{x^{k+1}\ln{x}}{k+1} - {\cfrac{x^{k+1}}{(k+1)^2}]^{1-\delta}}_{\epsilon}}$$

$$\large{= \cfrac{(1-\delta)^{k+1}\ln(1-\delta)}{k+1} - \cfrac{(1-\delta)^{k+1}}{(k+1)^2}}$$
$$\large{-\cfrac{{\epsilon}^{k+1}\ln{\epsilon}}{k+1} + \cfrac{{\epsilon}^{k+1}}{(k+1)^2}}$$

---------(9)

Thus, (7) becomes,

$$\large{\int\limits_{\epsilon}^{1-\delta}\ln{x}\ln{1-x}dx}$$

$$\large{= \ln{\epsilon}\sum_{k=1}^{\infty}\cfrac{{\epsilon}^{k+1}}{k(k+1)} - \sum_{k=1}^{\infty}\cfrac{{\epsilon}^{k+1}}{k(k+1)^2}}$$
$$\large{-\ln(1-\delta)\sum_{k=1}^{\infty}\cfrac{{1-\delta}^{k+1}}{k(k+1)} + \sum_{k=1}^{\infty}\cfrac{{1-\delta}^{k+1}}{k(k+1)^2}}$$ ---------- (10)

Thus,

$$\large{\int\limits_{\epsilon}^{1-\delta}\ln{x}\ln{1-x}dx}$$

$$\large{= \ln{\epsilon}*A(\epsilon) - B(\epsilon) - \ln(1-\delta)*A(1-\delta) + B(1-\delta)}$$ ----------- (11)

Where, for $\large{0<y<1}$,

$$\large{A(y) = \sum_{k=1}^{\infty}\cfrac{y^{k+1}}{k(k+1)}}$$ ------ (12)

$$\large{B(y) = \sum_{k=1}^{\infty}\cfrac{y^{k+1}}{k(k+1)^2}}$$ ------- (13)

Now,

For $\large{0<\epsilon<1}$;

$$\large{-\ln(1-\epsilon) = \epsilon + \cfrac{{\epsilon}^2}{2} + \cfrac{{\epsilon}^{3}}{3} + ...}$$

which is:

(i) $\large{>\epsilon}$

(ii) $\large{< \epsilon + {\epsilon}^2 + {\epsilon}^3 + ... = \cfrac{\epsilon}{1-\epsilon}}$

Thus,

$$\large{-\epsilon\ln{\epsilon} < (\ln{\epsilon})\ln(1-\epsilon) < \cfrac{-\epsilon\ln{\epsilon}}{1-\epsilon}}$$ -------- (14)

Now, since,

$$\large{\lim_{\epsilon \rightarrow 0^{+}}\epsilon\ln{\epsilon} = 0}$$

Thus,

$$\large{\lim_{\epsilon \rightarrow 0^{+}} (\ln{\epsilon})\ln{1-\epsilon} = 0}$$ ----- (15)

Also, for $\large{0<\epsilon<1}$,

$$\large{A(\epsilon) = \epsilon\sum_{k=1}^{\infty}\cfrac{{\epsilon}^k}{k} - \sum_{k=1}^{\infty}\cfrac{{\epsilon}^k}{k+1}}$$

$$\large{= -\epsilon\ln(1-\epsilon) + \ln{(1-\epsilon)} + \epsilon}$$

$$\large{= [\ln{(1-\epsilon)}](1-\epsilon) + \epsilon}$$ ------ (16)

Thus,

$$\large{\lim_{\epsilon \rightarrow 0^{+}}(\ln{\epsilon})A(\epsilon) = 0}$$ ------- (17)

Now, by Abel's theorem,

$$\large{\lim_{\delta \rightarrow 0^{+}} A(1-\delta) = \lim_{\delta \rightarrow 0^{+}} \sum_{k=1}^{\infty}\cfrac{{1-\delta}^{k+1}}{k(k+1)}}$$

$$\large{= \sum_{k=1}^{\infty}\cfrac{1}{k(k+1)} = 1}$$ --------- (18)

Thus,

$$\large{\lim_{\delta \rightarrow 0^{+}}(\ln(1-\delta)A(1-\delta)) = \ln{1} = 0}$$
-------(19)

Also,

$$\large{|B(\epsilon)| \le \epsilon \sum_{k=1}^{\infty}\cfrac{1}{k(k+1)^2}}$$ ------ (20)

Thus,

$$\large{\lim_{\epsilon \rightarrow 0^{+}} B(\epsilon) = 0}$$ ----- (21)

Again, by Abel's theorem,

$$\large{\lim_{\delta \rightarrow 0^{+}} B(1-\delta)}$$

$$\large{= \lim_{\delta \rightarrow 0^{+}} \sum_{k=1}^{\infty}\cfrac{(1-\delta)^{k+1}}{k(k+1)^2}}$$

$$\large{= \sum_{k=1}^{\infty}\cfrac{1}{k(k+1)^2}}$$

$$\large{= \sum_{k=1}^{\infty}[\cfrac{1}{k(k+1)} - \cfrac{1}{(k+1)^2}]}$$

$$\large{= \sum_{k=1}^{\infty}[\cfrac{1}{k} - \cfrac{1}{k+1}] - \sum_{k=1}^{\infty}[\cfrac{1}{(k+1)^2}]}$$

$$\large{= 1 - (\cfrac{{\pi}^2}{6} - 1)}$$

$$\large{= 2 - \cfrac{{\pi}^2}{6}}$$ --------- (22)

Thus, from (11),

$$\large{I = 2 - \cfrac{{\pi}^2}{6}}$$ -------- (23)

marissalovescats · Answer

This problem is not as cute as last nights :(

vishweshshrimali5 · Answer

Yeah. The problem has grown up.

vishweshshrimali5 · Answer

Its no more innocent :(

marissalovescats · Answer

LOL 

I did she the cutest problem today and I told you I'd let you know. 

It involved cats.

luigi0210 · Answer

That integral tho

vishweshshrimali5 · Answer

ERRATA: In the LATEX ERROR:
It is:
$$\large{y \ge 1}$$

anonymous · Answer

That's neat, thanks for sharing @vishweshshrimali5 :)

marissalovescats · Answer

Must of took forever to write out in the equation thing..

vishweshshrimali5 · Answer

It was my pleasure @iambatman . While I was solving this integral, I had a doubt whether I will get a correct answer as no solution was available. So, I spent over 4 hours solving this one. 

After I checked it with help of wolfram, I decided to share it with you all.

vishweshshrimali5 · Answer

@hartnn @ganeshie8 @Miracrown @ikram002p

vishweshshrimali5 · Answer

@mathmale @mathslover

vishweshshrimali5 · Answer

I would appreciate if you can provide an alternative method.

vishweshshrimali5 · Answer

As this was pretty lengthy. It took 7 pages of my notebook. :(

vishweshshrimali5 · Answer

@Zarkon

mathslover · Answer

Okay. lol keep it up. :P 
(Was too easy for me. so, I skipped the whole process,
and checked the answer. Answer is right :P )

vishweshshrimali5 · Answer

:P

vishweshshrimali5 · Answer

As, I had not read about Abel's theorem, so, that was the main part where I was stuck. I had to do a lot of research on internet and in various books of calculus.

But, my main intention is to let the students know that how various theorems can be applied for solving integrals. Abel's theorem helps in finding limit while Euler's formula for $\large{\cfrac{{\pi}^2}{6}}$ is a result of infinite series.

Though, various theorems may belong to different branches of maths they can be used to solve different questions from other branches.

Hope I was successful in my mission.

ikram002p · Answer

BANG lol
i cant read this my heard is full with 00011101010111010101 compination :P
i'll read it some other time , but seems like good work , ty for sharing :D

vishweshshrimali5 · Answer

It was my pleasure @ikram002p . Even, I will have to check this solution again for any latex errors.

mathslover · Answer

@Kainui will like to see it . :)

vishweshshrimali5 · Answer

@Kainui What do you think? Is there any mistake in this method ?

vishweshshrimali5 · Answer

I don't have any solution so just asking. :)

kainui · Answer

Well at first I gave you a medal because I like it but it will take me a little while to read through it. So far so good though, I just had to check what the power series of lnx was. I'll tell you when I'm done, hopefully soon! =)

vishweshshrimali5 · Answer

Thanks a lot @Kainui

kainui · Answer

I'm making sure I follow each step and understand exactly what you're doing. I'm really mystified though as to why you're doing what you're doing which is fascinating to me because everything seems like logical steps but I just don't know where it's going or how you thought to take any of these steps in the first place. I'm enjoying it! I'm at your 4th line right now to give you an idea of where I am.

anonymous · Answer

^ I thought I was the only one feeling like that haha.

kainui · Answer

@iambatman yeah inorite? wtf?

@vishweshshrimali5 what were you thinking in your mind or what was your goal as you were solving this? 

You just took the limit as n goes to infinity before line 7 and at first I thought, "Great, so what? Now it's 0 so there's nothi... OH! WOAH!" lol that's what I'm thinking right now, seems pretty cool let's see what's next!

kainui · Answer

Alright I'm also not very familiar with Abel's Theorem, I've heard about it but I've really not read much about it. I'm curious if you can explain it more. I like how it turns out to use euler's formula there for the sum of inverse squares. Very interesting!

I'm not sure how you thought to do this, but it definitely works. I'm happy to see something completely new like this. =)

@mathslover I'd like to see your "easy" way haha.

I'm looking for a better way right now. So far the only thing I've noticed is that it's an even function in the sense that the integral from 0 to 1/2 is the same as 1/2 to 1.

mathslover · Answer

lol Am a little bit tired... may be next year? :P

vishweshshrimali5 · Answer

@Kainui , I am happy to see that you were interested and fascinated by the solution.
Well, first of all for how I got those steps. Well, the truth is that I try to first draw a framework in my mind that how can I achieve the required solution. For example, here, I wanted to evaluate the integral but I knew from the very first look at the question that the function is discontinuous at limits. So, I knew that I had to use the limits concept. Next, I also knew that while evaluating the limits you must have some formulae in your mind and those included the expansion formulae. So, after wasting 6 pages, it appeared in my mind that why not give it a try by expansion formulae and there I got it. Next, I also knew that if I tried to use integration by parts, I am going to fail very badly, unless I got some general formula for nth integral. But, after giving it some try using that I understood its not THE one. So, I approached internet and searched for some pretty theorems and got to know about Abel's theorem. Actually, Abel's theorem is a very basic theorem to understand and apply. It basically helps to calculate limits easily provided that some conditions are fulfilled. One of these conditions was that the series must be converging and when I checked I got an expression, like "Hey! That's gonna work!". So, I tried using it and got the final answer. Initially, instead of using Abel's theorem, I was going to write the functions in expanded form and was going to use some very disturbing concepts which I had no idea of,but I decided against it.