1. Originally Posted by Isomorphism
$\int_0^\infty \frac{dx}{1+x^4}$ can be solved without complex numbers. I used only pure substitution and algebraic tricks

P.S: Unfortunately what I did only holds for this problem . It does not work for TPH's generalization.

While I was playing with $\int_0^\infty \frac{dx}{1+x^4}$, I found out an interesting way to prove $\int_0^\infty \frac{1 - x^2}{1+x^4}\, dx = 0$

With $x \to \frac1{x}$ substitution:

$\int_0^\infty \frac{dx}{1+x^4} = \int_0^\infty \frac{x^2 dx}{1+x^4}$

Its wonderful. Isnt it?

2. Here is the latest one:

$\boxed{I = \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)} \, dy \, dx}$

I invented it after seeing a recent problem

3. Part 2:
$\int^1_0\frac{x^4(1-x)^4}{1+x^2} dx$

This is a funny integral... not hard... but the answer is interesting
People who solve it can comment on the answer rather than showing the answer
We can do this one by just dividing. It's not that hard but the solution is cool.

$\int\left[x^{6}-4x^{5}+5x^{4}-4x^{2}-\frac{4}{x^{2}+1}+4\right]dx$

Which gives us the solution of $\frac{22}{7}-{\pi}$

4. Originally Posted by Isomorphism
Here is the latest update of "Definitely Integral"
Compute I:
$\boxed{I = \int_0^{\pi/4}\ln(\sin x) \cdot \ln(\cos x)\,dx}$

P.S:I think its quite hard
First note that: $
\int_0^{\tfrac{\pi }
{2}} {\ln \left[ {\sin \left( \theta \right)} \right] \cdot \ln \left[ {\cos \left( \theta \right)} \right]d\theta } = 2 \cdot \int_0^{\tfrac{\pi }
{4}} {\ln \left[ {\sin \left( \theta \right)} \right] \cdot \ln \left[ {\cos \left( \theta \right)} \right]d\theta }
$

Recall the identity: ${\rm B}\left( {x,y} \right) = 2 \cdot \int_0^{\tfrac{\pi }
{2}} {\sin ^{2x - 1} \left( \theta \right) \cdot \cos ^{2y - 1} \left( \theta \right)d\theta }
$

Thus: $
\left. {\frac{{\partial ^2 {\rm B}\left( {x,y} \right)}}
{{\partial x \cdot \partial y}}} \right|_{x = y = \tfrac{1}
{2}} = 8 \cdot \int_0^{\tfrac{\pi }
{2}} {\ln \left[ {\sin \left( \theta \right)} \right] \cdot \ln \left[ {\cos \left( \theta \right)} \right]d\theta }
$

Which can be computed using the identity: $
{\rm B}\left( {x,y} \right) = \frac{{\Gamma \left( x \right) \cdot \Gamma \left( y \right)}}
{{\Gamma \left( {x + y} \right)}}
$
(and some values of $
{\Gamma \left( x \right)}
$
; $
{\Gamma '\left( x \right)}
$
and $
{\Gamma ''\left( x \right)}
$
)

That's not a nice computation but it still works

5. Originally Posted by Isomorphism
Here is the latest one:

$\boxed{I = \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)} \, dy \, dx}$

I invented it after seeing a recent problem

$\int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)} ~ dy ~ dx$

$\int_0^1 \int_y^1 \frac{\ln (1 - y)}{y(y-1)} ~dx~dy$

$\int_0^1 x\frac{\ln (1 - y)}{y(y-1)} \bigg{|}_y^1~dy$

$\int_0^1 \frac{\ln (1-y)}{y(y-1)}-\frac{y\ln(1-y)}{y(y-1)} ~dy$

$- \int_0^1 \frac{\ln (1-y)}{y}~dy$

$\ln (1-y) = -\sum_{k=1}^{\infty} \frac{y^k}{k} = - \left ( y + \frac{y^2}{2}+\frac{y^3}{3}+\dots \right)$

$\frac{\ln (1-y)}{y} = -\sum_{k=1}^{\infty} \frac{y^{k-1}}{k} = - \left ( 1 + \frac{y}{2}+\frac{y^2}{3}+\dots \right )$

$- \int_0^1 \frac{\ln (1-y)}{y}~dy = \sum_{k=1}^{\infty} \frac{y^k}{k^2} \bigg{|}_0^1$

$\sum_{k=1}^{\infty} \frac{1}{k^2}= \frac{\pi^2}{6}$

6. Originally Posted by Isomorphism
Here is the latest update of "Definitely Integral"

Compute I:

$\boxed{I = \int_0^{\pi/4}\ln(\sin x) \cdot \ln(\cos x)\,dx}$
First, consider $\ln(1 + e^{2ix}) = \ln(2\cos x e^{ix})$

Comparing real parts you get $\ln(2\cos x) = \frac{\cos 2x}{1} - \frac{\cos 4x}{2} + \frac{\cos 6x}{6} - \cdots$

$\Rightarrow \ln(2 \sin x) = -\left(\frac{\cos 2x}{1} + \frac{\cos 4x}{2} + \frac{\cos 6x}{3} + \cdots \right)$

(I cannot rigorously justify these identities, but they seem to work at least for the range $0 \leq x \leq \frac{\pi}{2}$)

Note that $\int^{\frac{\pi}{4}}_{0} \ln(\sin x)\ln(\cos x) \;\mathrm{d}x = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x$

Also, if $m \not= n$, then $\int^{\frac{\pi}{2}}_{0} \cos 2mx \cos 2nx \; \mathrm{d}x = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \cos 2(m+n)x + \cos 2(m-n) x \; \mathrm{d}x = 0$

So $\int^{\frac{\pi}{2}}_{0} \ln(2\sin x)\ln(2\cos x) \; \mathrm{d}x = - \int^{\frac{\pi}{2}}_{0} \frac{\cos^{2}2x}{1^{2}} - \frac{\cos^{2}4x}{2^{2}} + \frac{\cos^{2}6x}{3^{2}} - \cdots \; \mathrm{d}x$

$\int^{\frac{\pi}{2}}_{0} \cos^{2}2nx \; \mathrm{d}x = \int^{\frac{\pi}{2}}_{0} \sin^{2}2nx \; \mathrm{d}x = \frac{\pi}{4}$

$\Rightarrow \int^{\frac{\pi}{2}}_{0} \ln(2\cos x)\ln(2\sin x) \; \mathrm{d}x = - \frac{\pi}{4}(\frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \cdots ) = - \frac{\pi^{3}}{48}$

$\ln(\sin x)\ln(\cos x) = \ln(2 \sin x)\ln(2 \cos x) - \ln2\left( \ln(\sin x) + \ln(\cos x)\right) - (\ln2)^{2}$

It is a well-known result that $\int^{\frac{\pi}{2}}_{0} \ln(\cos x) \; \mathrm{d}x = \int^{\frac{\pi}{2}}_{0} \ln(\sin x) \; \mathrm{d}x = -\frac{\pi}{2} \ln 2$

So $\int^{\frac{\pi}{2}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x = -\frac{\pi^{3}}{48} + \pi (\ln2)^{2} - \frac{\pi}{2}(\ln 2)^{2} = \frac{\pi }{2}(\ln 2)^{2} - \frac{\pi^{3}}{48}$

$\boxed{\Rightarrow \int^{\frac{\pi}{4}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x = \frac{\pi}{4} (\ln 2)^{2} - \frac{\pi^{3}}{96}}$

Awesome integral!

7. Originally Posted by wingless
$\int_0^1 x\frac{\ln (1 - y)}{y(y-1)} \bigg{|}_1^y~dy$
Good job but I think you mean $\bigg|_y^1$.

$- \int_0^1 \frac{\ln (1-y)}{y}~dy = \sum_{k=1}^{\infty} \frac{y^k}{k^2} \bigg{|}_0^1$
You are just missing an integral sign.

8. Originally Posted by Isomorphism
$\int_0^\infty \frac{dx}{1+x^4}$ can be solved without complex numbers. I used only pure substitution and algebraic tricks
Hmm, did you use the substitution $x = e^{u}$? That comes out fairly nicely, but there might be an easier way.

Let $I = \int^{\infty}_{0} \frac{1}{1 + x^{4}} \; \mathrm{d}x$

Let $x = e^{u}$

$I = \int^{\infty}_{-\infty} \frac{e^{u}}{1 + e^{4u}} \; \mathrm{d}x = \int^{\infty}_{-\infty} \frac{e^{-u}}{e^{2u} + e^{-2u}} \; \mathrm{d}x$

Let u = -t:

$I = \int^{\infty}_{-\infty} \frac{e^{t}}{e^{2t} + e^{-2t}} \;\mathrm{d}t$

$\Rightarrow 2I = \int^{\infty}_{-\infty} \frac{\cosh t}{\cosh 2t} \; \mathrm{d}t$

$= \int^{\infty}_{-\infty} \frac{\cosh t}{1 + 2\sinh^{2}t} \; \mathrm{d}t = \frac{\sqrt{2}}{2} \left[\tan^{-1}(\sinh t \cdot{\sqrt{2}})\right]^{\infty}_{-\infty} = \frac{\pi \sqrt{2}}{2}$

$\boxed{\Rightarrow \int^{\infty}_{0} \frac{1}{1 + x^{4}} \; \mathrm{d}x = \frac{\pi \sqrt{2}}{4}}$

9. Originally Posted by ThePerfectHacker
Good job but I think you mean $\bigg|_y^1$.
True, typo fixed.

Originally Posted by ThePerfectHacker
You are just missing an integral sign.
No, RHS is already integrated, otherwise it would be $\int_0^1 \sum_{k=1}^{\infty}\frac{y^{k-1}}{k}~dy$.

10. Here is another:

$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy$

that works out to $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2} }{6}$

11. Originally Posted by galactus
Here is another:
$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy$
$\frac{1}{1-xy} = \sum_{n=0}^{\infty}(xy)^n$.

Thus, $\int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^n y^n dx dy = \int_0^1 \sum_{n=0}^{\infty}\frac{1}{n+1}y^n dy = \sum_{n=0}^{\infty}\frac{1}{(n+1)^2} = \frac{\pi^2}{6}$.

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