1. I can 'explain' how the substitution works $x = \frac{1-t}{1+t}$ but actually it is not the explanation , i am going to change the integral to a form which is more intuitive .

$I = \int_0^{\pi/2} \frac{\sin(x)}{\sin(x) + \cos(x)} ~dx$

Intuition or experiences or your friend tell you to subsitute $x = \frac{\pi}{2} - y$ so we have

$I = \int_0^{\pi/2} \frac{\cos(y)}{\cos(y)+ \sin(y)} ~dy$

$= \int_0^{\pi/2} \frac{\cos(y) + \sin(y) }{\cos(y)+ \sin(y)} ~dy - \int_0^{\pi/2} \frac{\sin(y)}{\cos(y)+ \sin(y)} ~dy$

$= \frac{\pi}{2} - I$ so $I = \frac{\pi}{4}$

Here is another one :

$I = \int_0^{\pi} \frac{x\sin(x)}{1+ \cos^2(x) } ~dx$

Again , they tell us to subsitute $x = \pi - y$

$I = \int_0^{\pi} \frac{(\pi - y)\sin(y) }{1 + \cos^2(y) }~dy$

$= \pi \int_0^{\pi} \frac{\sin(y)}{1 + \cos^2(y) }~dy - I$

$= \pi \left[ - \tan^{-1}[\cos(y)] \right]_0^{\pi} - I$

$= \frac{\pi^2}{2} - I$ so

$I = \frac{\pi^2}{4}$

The tricks of the above two integrals are very similar , what we do is to 'reflect' the integrand to obtain the answers . But we can't explain why one could think of , our intuition always cannot be explained .

This one is very similar .

Sub. $x = \tan(y)$

$dx = \sec^2(y) ~dy$ we have

$I = \int_0^{\pi/4} \ln[1 + \tan(y) ] ~dy$

Again , sub $y = \frac{\pi}{4} - z$ You may find that it is equivalent to sub $x = \frac{1-t}{1+t}$ at the begining as $\tan( \frac{\pi}{4} - \theta ) = \frac{1 - \tan(\theta) }{1 + \tan(\theta) }$

Therefore , $I = \int_0^{\pi/4} \ln{ \left( 1 + \frac{1 - \tan(z) }{1 + \tan(z) \right) }~dz$

$= \int_0^{\pi/4} \ln(2) ~dz - I$

$= \frac{\pi \ln(2)}{4} - I$ so

$I = \frac{\pi \ln(2)}{8}$

You see , these three integrals are not different at all so if you are asking how this works , let me ask you how this also works when applied in the first two integrals .

2. Compute $\displaystyle \int_0^1 \frac{\log^ax}{x^2+1} dx$, for $a\geq0$.

3. The substitutions exploit the symmetries of the integrals. Is that what you're getting at?

4. Originally Posted by chiph588@
Compute $\displaystyle \int_0^1 \frac{\log^ax}{x^2+1} dx$, for $a\geq0$.
Spoiler:

Let

$\displaystyle I(a)=\int_0^1\frac{\log^a(x)}{1+x^2}\text{ }dx$

Note then that for $x\in[0,1]$ we have that

$\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}$

So that

\displaystyle \begin{aligned}I(a) &= \int_0^1\log^a(x)\sum_{n=0}^{\infty}(-1)^n x^{2n}\text{ }dx\\ &= \sum_{n=0}^{\infty}(-1)^n \int_0^1 \log^a(x) x^{2n}\text{ }dx\end{aligned}

But, let $y=-log(x)$ to get that

$\displaystyle \int_0^1\log^a(x)x^{2n}\text{ }dx = (-1)^a\int_{0}^{\infty}y^a e^{-(2n+1)y}\text{ }dy$

Letting $(2n+1)y=z$ gives that

\displaystyle \begin{aligned}(-1)^a\int_0^{\infty}y^a e^{-(2n+1)y}\text{ }dy &= \frac{(-1)^a}{(2n+1)^{a+1}}\int_0^{\infty}z^a e^{-z}\text{ }dz\\ &= \frac{(-1)^a \Gamma(a+1)}{(2n+1)^{a+1}}\end{aligned}

Thus,

\displaystyle \begin{aligned}I(a) &= \sum_{n=0}^{\infty}(-1)^n \int_0^1 \log^a(x) x^{2n}\text{ }dx\\ &=(-1)^a \Gamma(a+1)\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^{a+1}}\\ &= (-1)^a\Gamma(a+1)\beta(a+1)\end{aligned}

Which is now clear since letting $-\log(x)=y$ in our original integral we see that

$\displaystyle \int_0^1\frac{\log^a(x)}{1+x^2}\text{ }dx=\frac{(-1)^a}{2}\int_0^{\infty}\frac{y^a}{\cosh(x)}\text{ }dx$

5. How about showing that $\int^{1}_{0} \frac{\ln(x^2-x+1)}{x-1} \ dx = \frac{\pi^{2}}{18}$ ? I know of one approach, but I'm sure there are others.

6. I find one , first put $x = 1 -t$ we have

$\int_0^1 \frac{\ln(x^2 - x + 1 ) }{x-1}~dx$

$= - \int_0^1 \frac{\ln(t^2 - t + 1 )}{ t}~dt$

$= \int_0^1 \frac{\ln(t+1)}{t}~dt - \int_0^1 \frac{\ln(t^3+1)t^2}{t^3}~dt$

Sub. $t^3 = u$ in the second integral ( Since the limit at zero of the integrand exists , i think i can make the substitution without changing to $\displaystyle \lim_{a\to 0 } \int_a^1$ something . )

$= \int_0^1 \frac{\ln(t+1)}{t}~dt - \frac{1}{3} \int_0^1 \frac{\ln(u+1)}{u}~du$

$= ( 1 - \frac{1}{3} ) \int_0^1 \frac{\ln(t+1)}{t}~dt = \frac{2}{3} ~ \frac{\pi^2}{12}$

$= \frac{\pi^2}{18}$

7. You always find a better way.

I let $I(a) = \int^{1}_{0} \frac{\ln\big(a(x^{2}-x)+1\big)}{x-1} \ dx$, differentiated with respect to $a$, integrated, and then used the fact that $I(0)=0.$

EDIT: It may be be more correct to let $I(a) = \int^{1}_{0} \frac{\ln\Big(\frac{1}{a}(x^2-x)+1\Big)}{x-1} \ dx$ and use the fact that $I(\infty) = 0$.

Originally Posted by simplependulum
I find one , first put $x = 1 -t$ we have

$\int_0^1 \frac{\ln(x^2 - x + 1 ) }{x-1}~dx$

$= - \int_0^1 \frac{\ln(t^2 - t + 1 )}{ t}~dt$

$= \int_0^1 \frac{\ln(t+1)}{t}~dt - \int_0^1 \frac{\ln(t^3+1)t^2}{t^3}~dt$

Sub. $t^3 = u$ in the second integral ( Since the limit at zero of the integrand exists , i think i can make the substitution without changing to $\displaystyle \lim_{a\to 0 } \int_a^1$ something . )

$= \int_0^1 \frac{\ln(t+1)}{t}~dt - \frac{1}{3} \int_0^1 \frac{\ln(u+1)}{u}~du$

$= ( 1 - \frac{1}{3} ) \int_0^1 \frac{\ln(t+1)}{t}~dt = \frac{2}{3} ~ \frac{\pi^2}{12}$

$= \frac{\pi^2}{18}$

8. Here's one I just came up with. Compute $\displaystyle \sum_{k=0}^\infty (-1)^k \frac{2k+3}{(3k+4)(3k+5)}$.

9. And interesting corollary of this is that $\displaystyle \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}=\sum_{n=1}^{ \infty}\frac{(n!)^2}{n^2 (2n)!}=\frac{\pi^2}{18}$.

10. Originally Posted by Drexel28
And interesting corollary of this is that $\displaystyle \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}=\sum_{n=1}^{ \infty}\frac{(n!)^2}{n^2 (2n)!}=\frac{\pi^2}{18}$.
I'm assuming you got to this by a substitution of some kind and then expanded into series. If so, what was your substitution?

11. Originally Posted by chiph588@
I'm assuming you got to this by a substitution of some kind and then expanded into series. If so, what was your substitution?
No substitution,

\displaystyle \begin{aligned}\frac{\pi^2}{18} &= \int_0^1\frac{\log\left(1-x(x-1)\right)}{x-1}\text{ }dx\\ &= \int_0^1\frac{1}{x-1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n(1-x)^n\\ &= \sum_{n=1}^{\infty}\frac{1}{n}\int_0^1 x^n(1-x)^{n-1}\text{ }dx\\ &= \sum_{n=1}^{\infty}\frac{B(n+1,n)}{n}\\ &= \sum_{n=1}^{\infty}\frac{\Gamma(n+1)\Gamma(n)}{n\G amma(2n+1)}\\ &= \sum_{n=1}^{\infty}\frac{(n!)^2}{n^2 (2n)!}\end{aligned}

I believe that every lover of series and integrals should do the following four at least once in their life

$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^k(2n)!},\quad k=0,1,2,4$

and compute

$\displaystyle \lim_{k\to\infty}\sum_{n=1}^{\infty}\frac{(n!)^2}{ n^k (2n)!}$

12. Originally Posted by chiph588@
Here's one I just came up with. Compute $\displaystyle \sum_{k=0}^\infty (-1)^k \frac{2k+3}{(3k+4)(3k+5)}$.
Not to be a problem spoiler, but is there some trick to compute this? There is a really, really ugly way. But I don't feel like carrying that out haha.

13. Originally Posted by Drexel28
Not to be a problem spoiler, but is there some trick to compute this? There is a really, really ugly way. But I don't feel like carrying that out haha.
Let's start off with a small hint and see where that gets you. I solved this one by accident when trying to solve the last integral Random Variable posted.

14. Originally Posted by Drexel28
$\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^k(2n)!},\quad k=0$
$\displaystyle S_0 = \sum_{n=0}^\infty \frac{n!}{(2n)!} = \sum_{n=0}^\infty \frac1{\binom{2n}{n}n!}$

Now notice $\displaystyle \binom{2n}{n}^{-1} = (2n+1)\int_0^1t^n(1-t)^n dt$. Thus $\displaystyle S_0 = \int_0^1 \sum_{n=0}^\infty \frac{(2n+1)(t-t^2)^n}{n!} dt = \int_0^1 (-2t^2+2t+1)e^{t-t^2} dt$.

Unfortunately this has no closed integral but can be whittled down to $\displaystyle S_0 = 1+e^{1/4}\int_0^{1/2}e^{-t^2} dt$.

15. Originally Posted by chiph588@
$\displaystyle S_0 = \sum_{n=0}^\infty \frac{n!}{(2n)!} = \sum_{n=0}^\infty \frac1{\binom{2n}{n}n!}$

Now notice $\displaystyle \binom{2n}{n}^{-1} = (2n+1)\int_0^1t^n(1-t)^n dt$. Thus $\displaystyle S_0 = \int_0^1 \sum_{n=0}^\infty \frac{(2n+1)(t-t^2)^n}{n!} dt = \int_0^1 (-2t^2+2t+1)e^{t-t^2} dt$.

Unfortunately this has no closed integral but can be whittled down a bit to $\displaystyle S_0 = e^{1/4}\int_0^{1/2}e^{-t^2} dt$.
Spoiler:

Let

$\displaystyle S=\sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}$

Notice then that,

\displaystyle \begin{aligned} S &= \sum_{n=1}^{\infty}\frac{n n! (n-1)!}{(2n)!}\\ &= \sum_{n=1}^{\infty}\frac{n \Gamma(n+1)\Gamma(n)}{\Gamma(2n+1)}\\ &= \sum_{n=1}^{\infty}n B(n+1,n)\\ &= \sum_{n=1}^{\infty}n\int_0^1 x^{n-1}(1-x)^n\text{ }dx\\ &= \int_0^1 \frac{1}{x}\sum_{n=1}^{\infty}n\left(x(1-x)\right)^n\\ &= \int_0^1\frac{1-x}{(1-x+x^2)^2}\text{ }dx\\ &= \frac{1}{27}\left(9+2\sqrt{3}\pi\right)\end{aligne d}

I'll leave you to verify this last integral.

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