1. Here is another one, (if you are tired of them blame my dad for having a CRC math of mathematical formulas and tables)

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

$\displaystyle =\int_0^{1}\ln(x)\sum_{n=0}^{\infty}x^{2n}~dx$

$\displaystyle =\int_0^{1}\sum_{n=0}^{\infty}\ln(x)x^{2n}~dx$

$\displaystyle =\sum_{n=0}^{\infty}\int_0^{1}x^{2n}\ln(x)~dx$

$\displaystyle =\sum_{n=0}^{\infty}\bigg[\left(\frac{x\ln(x)}{2y+1}-\frac{x}{(2y+1)^2}\right)x^{2y}\bigg]\bigg|_0^{1}$

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

$\displaystyle =\frac{-\pi^2}{8}$

From this we can show that

$\displaystyle \int_0^{1}\frac{{\rm{arctanh}}(x)}{x}~dx=\frac{\pi ^2}{8}$

How we do that is seeing that

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

$\displaystyle =-\int_0^{1}\int_1^x\frac{dy}{y(1-x^2)}~dy~dx$

$\displaystyle =-\int_0^1\int_0^y\frac{dx}{y(1-x^2)}~dt$

$\displaystyle =-\int_0^{1}\frac{1}{y}\bigg[{\rm{arctanh}}(x)\bigg|_0^{y}~dy$

$\displaystyle =-\int_0^{1}\frac{{\rm{arctanh}}(y)}{y}~dy$

$\displaystyle \Rightarrow{-\int_0^{1}\frac{{\rm{arctanh}}(y)}{y}~dy=\int_0^1\ frac{\ln(x)}{1-x^2}~dx}$

$\displaystyle \Rightarrow{-\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy=\frac{-\pi^2}{8}}$

$\displaystyle \Rightarrow\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy= \frac{\pi^2}{8}\quad\blacksquare$

Is there another way to do this last integral? A hint please.

2. Ths one has taken me like two hours to do with all the calculations.

Firstly we must establish something

$\displaystyle \frac{1}{6}\int_0^{\infty}{y^3e^{-yx}~dy}=\frac{1}{x^4}$

Proof: By integration by Parts

$\displaystyle \int{y^3e^{-yx}}~dy$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\int{y^2e^{-yx}}~dy$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\int{ye^{-yx}~dy}\bigg]$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\bigg[\frac{-1}{x}ye^{-yx}+\frac{1}{x}\int{e^{-yx}}~dy\bigg]\bigg]$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\bigg[\frac{-1}{x}ye^{-yx}+\frac{-1}{x^2}e^{-yx}\bigg]\bigg]$

$\displaystyle =\frac{-(x^3y^3+3x^2y^2+6xy+6)e^{-yx}}{x^4}$

Now if we define

$\displaystyle F(x)=\frac{-(x^3y^3+3x^2y^2+6xy+6)e^{-yx}}{x^4}$

Then

$\displaystyle \int_0^{\infty}y^3e^{-yx}~dy=F(\infty)-F(0)$

Now seeing that

$\displaystyle \lim_{x\to\infty}F(x)=0$

Due to the overpowering effect of $\displaystyle e^{-yx}$

And that

$\displaystyle F(0)=\frac{-6}{x^4}$

Then

$\displaystyle \int_0^{\infty}y^3e^{-yx}~dy$

$\displaystyle =F(\infty)-F(0)$

$\displaystyle =0-\frac{-6}{x^4}$

$\displaystyle =\frac{6}{x^4}$

$\displaystyle \Rightarrow\frac{1}{6}\int_0^{\infty}y^3e^{-yx}~dy$

$\displaystyle =\frac{1}{6}\cdot\frac{6}{x^4}$

$\displaystyle =\frac{1}{x^4}\quad\blacksquare$

So now we can get to the actual integral

$\displaystyle \int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx$

My first intuition was to do as Krizalid said with the other and transform it into a function of cosines, but I found the working with the functions to be to cumbersome so I kept it as is and used the above knowledge to transform it as follows

$\displaystyle \int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx$

$\displaystyle \int_0^{\infty}\sin^4p(x)\frac{1}{6}\int_0^{\infty }y^3e^{-yx}~dy~dx$

$\displaystyle =\frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dy~dx$

Now noting that we are integrating over a rectangular region the ubiquitous Fubini's Theorem applies enabling us to switch the integration order scott free. Thus

$\displaystyle \frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dy~dx$

$\displaystyle =\frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dx~dy$

Now here comes something horrible we must calculate

$\displaystyle \int{y^3e^{-yx}\sin^4(px)}~dx$

It took me an hour alone to do this calculation and it would take an eternity to type,so instead I will include the entire calculation of an integral of the same concept so those who read the others and did not see how to do it may follow along.

For the sake of alleviating calculations we know that

$\displaystyle \int{y^3e^{-yx}\sin(px)}~dx$

$\displaystyle =y^3\int{e^{-yx}\sin(px)}~dx$

So by Integration by Parts

$\displaystyle \int{e^{-yx}\sin(px)}$

$\displaystyle =\frac{-1}{y}\sin(px)e^{-yx}+\frac{p}{y}\int{e^{-yx}\cos(px)}~dx$

$\displaystyle =\frac{-1}{y}\sin(px)e^{-yx}+\frac{p}{y}\bigg[\frac{-1}{y}\cos(px)e^{-yx}+\frac{p}{y}\int{e^{-yx}\sin(px)~dx}\bigg]$

So if we expand this we get

$\displaystyle \int{e^{-yx}\sin(px)}~dx=\frac{-1}{y}\sin(px)e^{-yx}-\frac{p}{y^2}\cos(px)e^{-yx}+\frac{p^2}{y^2}\int{e^{-yx}\sin(px)}~dx$

Now you notice that if we add the last term and factor we get

$\displaystyle \left(1+\frac{p^2}{y^2}\right)\int{e^{-yx}\sin(px)}~dx=\frac{-e^{-yx}}{y}\left(\sin(px)+\frac{p}{y}\cos(px)\right)$

$\displaystyle \Rightarrow\int{e^{-yx}\sin(px)}~dx=\frac{\frac{-e^{-yx}}{y}\left(\sin(px)+\frac{p}{y}\cos(px)\right)}{ 1+\frac{p^2}{y^2}}$

$\displaystyle =\frac{-ye^{-yx}\left(\sin(px)+\frac{p}{y}\cos(px)\right)}{y^2+ p^2}$

So if you would actually do the calculation you would see that

$\displaystyle \int{y^3e^{-yx}\sin^4(px)}~dx$

$\displaystyle =y^3\bigg[\left(\frac{-p\sin^3(px)}{y^2+16p^2}-\frac{24p^3\sin(px)}{(y^2+4p^2)(y^2+16p^2)}\right)$$\displaystyle \cdot\cos(px)-\frac{\sin^4(px)y}{y^2+16p^2}-\frac{12p^2\sin^2(px)y}{(y^2+4p^2)(y^2+16p^2)}-\frac{24p^4}{y(y^2+4p^2)(y^2+16p^2)}\bigg]e^{-yx} And if you think that is bad you should have seen it prior to me letting my TI-89 clean it up for me. So if we let \displaystyle F(x)=y^3\bigg[\left(\frac{-p\sin^3(px)}{y^2+16p^2}-\frac{24p^3\sin(px)}{(y^2+4p^2)(y^2+16p^2)}\right)$$\displaystyle \cdot\cos(px)-\frac{\sin^4(px)y}{y^2+16p^2}-\frac{12p^2\sin^2(px)y}{(y^2+4p^2)(y^2+16p^2)}-\frac{24p^4}{y(y^2+4p^2)(y^2+16p^2)}\bigg]e^{-yx}$

Then we can see that

$\displaystyle \int_0^{\infty}y^3e^{-yx}\sin(px)~dx=F(\infty)-F(0)$

From there we can see that

$\displaystyle \lim_{x\to\infty}F(x)=0$

Due to the overpowering effects of $\displaystyle e^{-yx}$

And

$\displaystyle F(0)=\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}$

Then we see that

$\displaystyle \int_0^{\infty}y^3e^{-yx}\sin(px)~dx$

$\displaystyle =-\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]$

So now we must calculate

$\displaystyle -\int\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]~dy$

So if we look at them seperately we get

$\displaystyle \int\frac{8p^4}{y^2+4p^2}$

$\displaystyle \underbrace{=}_{y=2p\tan\theta}=4p^3\int\frac{d\th eta}{\theta^2+1}$

$\displaystyle \Rightarrow\int\frac{8p^4}{y^2+4p^2}$

$\displaystyle =4p^3\tan\left(\frac{y}{2p}\right)=R(y)$

And similarly

$\displaystyle \int\frac{32p^4}{y^2+16p^2}$

$\displaystyle \underbrace{=}_{y=4p\tan(\theta)}8p^3\int\frac{d\t heta}{\theta^2+1}$

$\displaystyle \Rightarrow\int\frac{4p^4}{y^2+16p^2}~dy$

$\displaystyle =8p^3\arctan\left(\frac{y}{4p}\right)=G(y)$

So

$\displaystyle -\int_0^{\infty}\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]~dy$

$\displaystyle =-\bigg[R(\infty)+G(\infty)-\left(G(0)+R(0)\right)\bigg]$

$\displaystyle =2p^3\pi$

$\displaystyle \Rightarrow\frac{1}{6}\int_0^{\infty}\int_0^{\inft y}y^3e^{-yx}\sin(px)$

$\displaystyle =\frac{p^3\pi}{3}$

So

$\displaystyle \boxed{\int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx=\f rac{p^3\pi}{3}}\quad\blacksquare$

So to match what Galactus's website said

$\displaystyle \int_0^{\infty}\frac{\sin^4(x)}{x^4}~dx=\frac{\pi} {3}$

Hope that was long enough for you . There has to be a better way to do it, yeah?

3. Originally Posted by Mathstud28
Ths one has taken me like two hours to do with all the calculations.

Firstly we must establish something

$\displaystyle \frac{1}{6}\int_0^{\infty}{y^3e^{-yx}~dy}=\frac{1}{x^4}$

Proof: By integration by Parts

$\displaystyle \int{y^3e^{-yx}}~dy$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\int{y^2e^{-yx}}~dy$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\int{ye^{-yx}~dy}\bigg]$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\bigg[\frac{-1}{x}ye^{-yx}+\frac{1}{x}\int{e^{-yx}}~dy\bigg]\bigg]$

$\displaystyle =\frac{-1}{x}y^3e^{-yx}+\frac{3}{x}\bigg[\frac{-1}{x}y^2e^{-yx}+\frac{2}{x}\bigg[\frac{-1}{x}ye^{-yx}+\frac{-1}{x^2}e^{-yx}\bigg]\bigg]$

$\displaystyle =\frac{-(x^3y^3+3x^2y^2+6xy+6)e^{-yx}}{x^4}$

Now if we define

$\displaystyle F(x)=\frac{-(x^3y^3+3x^2y^2+6xy+6)e^{-yx}}{x^4}$

Then

$\displaystyle \int_0^{\infty}y^3e^{-yx}~dy=F(\infty)-F(0)$

Now seeing that

$\displaystyle \lim_{x\to\infty}F(x)=0$

Due to the overpowering effect of $\displaystyle e^{-yx}$

And that

$\displaystyle F(0)=\frac{-6}{x^4}$

Then

$\displaystyle \int_0^{\infty}y^3e^{-yx}~dy$

$\displaystyle =F(\infty)-F(0)$

$\displaystyle =0-\frac{-6}{x^4}$

$\displaystyle =\frac{6}{x^4}$

$\displaystyle \Rightarrow\frac{1}{6}\int_0^{\infty}y^3e^{-yx}~dy$

$\displaystyle =\frac{1}{6}\cdot\frac{6}{x^4}$

$\displaystyle =\frac{1}{x^4}\quad\blacksquare$

So now we can get to the actual integral

$\displaystyle \int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx$

My first intuition was to do as Krizalid said with the other and transform it into a function of cosines, but I found the working with the functions to be to cumbersome so I kept it as is and used the above knowledge to transform it as follows

$\displaystyle \int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx$

$\displaystyle \int_0^{\infty}\sin^4p(x)\frac{1}{6}\int_0^{\infty }y^3e^{-yx}~dy~dx$

$\displaystyle =\frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dy~dx$

Now noting that we are integrating over a rectangular region the ubiquitous Fubini's Theorem applies enabling us to switch the integration order scott free. Thus

$\displaystyle \frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dy~dx$

$\displaystyle =\frac{1}{6}\int_0^{\infty}\int_0^{\infty}y^3e^{-yx}\sin^4(px)~dx~dy$

Now here comes something horrible we must calculate

$\displaystyle \int{y^3e^{-yx}\sin^4(px)}~dx$

It took me an hour alone to do this calculation and it would take an eternity to type,so instead I will include the entire calculation of an integral of the same concept so those who read the others and did not see how to do it may follow along.

For the sake of alleviating calculations we know that

$\displaystyle \int{y^3e^{-yx}\sin(px)}~dx$

$\displaystyle =y^3\int{e^{-yx}\sin(px)}~dx$

So by Integration by Parts

$\displaystyle \int{e^{-yx}\sin(px)}$

$\displaystyle =\frac{-1}{y}\sin(px)e^{-yx}+\frac{p}{y}\int{e^{-yx}\cos(px)}~dx$

$\displaystyle =\frac{-1}{y}\sin(px)e^{-yx}+\frac{p}{y}\bigg[\frac{-1}{y}\cos(px)e^{-yx}+\frac{p}{y}\int{e^{-yx}\sin(px)~dx}\bigg]$

So if we expand this we get

$\displaystyle \int{e^{-yx}\sin(px)}~dx=\frac{-1}{y}\sin(px)e^{-yx}-\frac{p}{y^2}\cos(px)e^{-yx}+\frac{p^2}{y^2}\int{e^{-yx}\sin(px)}~dx$

Now you notice that if we add the last term and factor we get

$\displaystyle \left(1+\frac{p^2}{y^2}\right)\int{e^{-yx}\sin(px)}~dx=\frac{-e^{-yx}}{y}\left(\sin(px)+\frac{p}{y}\cos(px)\right)$

$\displaystyle \Rightarrow\int{e^{-yx}\sin(px)}~dx=\frac{\frac{-e^{-yx}}{y}\left(\sin(px)+\frac{p}{y}\cos(px)\right)}{ 1+\frac{p^2}{y^2}}$

$\displaystyle =\frac{-ye^{-yx}\left(\sin(px)+\frac{p}{y}\cos(px)\right)}{y^2+ p^2}$

So if you would actually do the calculation you would see that

$\displaystyle \int{y^3e^{-yx}\sin^4(px)}~dx$

$\displaystyle =y^3\bigg[\left(\frac{-p\sin^3(px)}{y^2+16p^2}-\frac{24p^3\sin(px)}{(y^2+4p^2)(y^2+16p^2)}\right)$$\displaystyle \cdot\cos(px)-\frac{\sin^4(px)y}{y^2+16p^2}-\frac{12p^2\sin^2(px)y}{(y^2+4p^2)(y^2+16p^2)}-\frac{24p^4}{y(y^2+4p^2)(y^2+16p^2)}\bigg]e^{-yx} And if you think that is bad you should have seen it prior to me letting my TI-89 clean it up for me. So if we let \displaystyle F(x)=y^3\bigg[\left(\frac{-p\sin^3(px)}{y^2+16p^2}-\frac{24p^3\sin(px)}{(y^2+4p^2)(y^2+16p^2)}\right)$$\displaystyle \cdot\cos(px)-\frac{\sin^4(px)y}{y^2+16p^2}-\frac{12p^2\sin^2(px)y}{(y^2+4p^2)(y^2+16p^2)}-\frac{24p^4}{y(y^2+4p^2)(y^2+16p^2)}\bigg]e^{-yx}$

Then we can see that

$\displaystyle \int_0^{\infty}y^3e^{-yx}\sin(px)~dx=F(\infty)-F(0)$

From there we can see that

$\displaystyle \lim_{x\to\infty}F(x)=0$

Due to the overpowering effects of $\displaystyle e^{-yx}$

And

$\displaystyle F(0)=\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}$

Then we see that

$\displaystyle \int_0^{\infty}y^3e^{-yx}\sin(px)~dx$

$\displaystyle =-\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]$

So now we must calculate

$\displaystyle -\int\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]~dy$

So if we look at them seperately we get

$\displaystyle \int\frac{8p^4}{y^2+4p^2}$

$\displaystyle \underbrace{=}_{y=2p\tan\theta}=4p^3\int\frac{d\th eta}{\theta^2+1}$

$\displaystyle \Rightarrow\int\frac{8p^4}{y^2+4p^2}$

$\displaystyle =4p^3\tan\left(\frac{y}{2p}\right)=R(y)$

And similarly

$\displaystyle \int\frac{32p^4}{y^2+16p^2}$

$\displaystyle \underbrace{=}_{y=4p\tan(\theta)}8p^3\int\frac{d\t heta}{\theta^2+1}$

$\displaystyle \Rightarrow\int\frac{4p^4}{y^2+16p^2}~dy$

$\displaystyle =8p^3\arctan\left(\frac{y}{4p}\right)=G(y)$

So

$\displaystyle -\int_0^{\infty}\bigg[\frac{8p^4}{y^2+4p^2}-\frac{32p^4}{y^2+16p^2}\bigg]~dy$

$\displaystyle =-\bigg[R(\infty)+G(\infty)-\left(G(0)+R(0)\right)\bigg]$

$\displaystyle =2p^3\pi$

$\displaystyle \Rightarrow\frac{1}{6}\int_0^{\infty}\int_0^{\inft y}y^3e^{-yx}\sin(px)$

$\displaystyle =\frac{p^3\pi}{3}$

So

$\displaystyle \boxed{\int_0^{\infty}\frac{\sin^4(px)}{x^4}~dx=\f rac{p^3\pi}{3}}\quad\blacksquare$

So to match what Galactus's website said

$\displaystyle \int_0^{\infty}\frac{\sin^4(x)}{x^4}~dx=\frac{\pi} {3}$

Hope that was long enough for you . There has to be a better way to do it, yeah?
Ha. That was way too long... Great work though! [as usual]

--Chris

4. Originally Posted by Chris L T521
Ha. That was way too long... Great work though! [as usual]

--Chris
There were two shocking things about this, firstly it actually took 5 times longer to do than to type it (believe it or not) and secondly I did not make one LaTeX mistake the whole time...its a miracle

5. This is what I am currently learning so I think I will apply it

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}~dx$

Now if we let $\displaystyle \sqrt{a}x=z$

Then we get

$\displaystyle \frac{1}{a}\int_{-\infty}^{\infty}e^{-z^2}$

Now let us consider $\displaystyle I=\int_{-\infty}^{\infty}e^{-z^2}~dz$

and $\displaystyle J=\int_{-\infty}^{\infty}e^{-y^2}~dy$

Now consider that $\displaystyle J=I\Rightarrow{J\cdot{I}=I^2=J^2}$

So we may do the following

$\displaystyle I\cdot{J}=\int_{-\infty}^{\infty}e^{-x^2}~dx\cdot\int_0^{\infty}e^{-y^2}~dy$

$\displaystyle =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(y^2+z^2)}~dy~dz$

$\displaystyle =I^2$

Now I think this is correct but, I think thatt we may multiply them for although they aer numerically equivalent in terms of variables they are still constants? (Can someone back me up on this)

So if we convert to polar coordinates we get

$\displaystyle 2\pi\int_{0}^{\infty}re^{-r^2}~dr$

The extra r was given because of the Jacobian when converting to polar coordinates

So

$\displaystyle \int_0^{\infty}re^{-r^2}~dr$

$\displaystyle =\frac{1}{2}$

so

$\displaystyle 2\pi\int_0^{\infty}re^{-r^2}~dr$

$\displaystyle =\pi$

But remembering that we actually have

$\displaystyle \frac{2\pi}{a}\int_0^{\infty}re^{-r^2}~dr$

$\displaystyle \Rightarrow{I^2=\frac{\pi}{a}}$

$\displaystyle \Rightarrow{I}=\int_0^{\infty}e^{-ax^2}~dx=\sqrt{\frac{\pi}{a}}\quad\blacksquare$

EDIT:

I also just noticed that if you called

$\displaystyle I=\int_{-\infty}^{\infty}e^{-ax^2}~dx$

Then

$\displaystyle I^2=\int_{-\infty}^{\infty}\int_0={-\infty}^{\infty}e^{-a(x^2+y^2)}~dy~dx$

In other words there was no need for a sub

EDIT: Another typo

6. Originally Posted by Mathstud28
EDIT:

I also just noticed that if you called

$\displaystyle I=\int_0^{\infty}e^{-ax^2}~dx$

Then

$\displaystyle I^2=\int_0^{\infty}\int_0^{\infty}e^{-a(x^2+y^2)}~dy~dx$

In other words there was no need for a sub

Didn't you know about that trick prior to this revelation of yours??

--Chris

7. Originally Posted by Mathstud28
EDIT:

I also just noticed that if you called

$\displaystyle I=\int_{-\infty}^{\infty}e^{-ax^2}~dx$

Then

$\displaystyle I^2=\int_{-\infty}^{\infty}\int_{{\color{red}-\infty}}^{\infty}e^{-a(x^2+y^2)}~dy~dx$

In other words there was no need for a sub

EDIT: Another typo
Another typo...

Plus, if you were to integrate $\displaystyle \int_0^{\infty}e^{-ax^2}\,dx$, you should get $\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$...

Thus, $\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$

--Chris

8. Originally Posted by Chris L T521

Didn't you know about that trick prior to this revelation of yours??

--Chris
I don't think so, but I must have seen it..I am not that clever

Originally Posted by Chris L T521
Another typo...

Plus, if you were to integrate $\displaystyle \int_0^{\infty}e^{-ax^2}\,dx$, you should get $\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$...

Thus, $\displaystyle \int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$

--Chris
Oops ...and yes this would be because since

$\displaystyle \int{e^{-x^2}~dx}=$

$\displaystyle \int\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$

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

We can see that it is even

9. $\displaystyle \int_0^1\frac{\ln(1-x)}{x}~dx$

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

$\displaystyle \underbrace{=}_{\text{change of IO}}\int_0^1\int_1^y\frac{dx}{x(1-y)}~dy$

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

Now I have problaby already done this one...but....so what...

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

$\displaystyle =\int_0^1\ln(y)\sum_{n=0}^{\infty}y^n~dy$

$\displaystyle =-\sum_{n=0}^{\infty}\int_0^1\ln(y)y^n~dy$

Now I know I have used it before but I don't think I have "proved" it.

$\displaystyle \int_0^1\ln(y)y^n~dy=\frac{-1}{(n+1)^2}$

By Integration by Parts

$\displaystyle \int\ln(y)y^n~dy$

$\displaystyle =\frac{\ln(y)y^{n+1}}{n+1}-\frac{1}{n+1}\int\frac{y^{n+1}}{y}~dy$

$\displaystyle =\frac{\ln(y)y^{n+1}}{n+1}-\frac{y^{n+1}}{(n+1)^2}=F(y)$

Now we see that

$\displaystyle F(1)=\frac{-1}{(n+1)^2}$

But what about $\displaystyle F(0)$?

Well the second term obviously drops out, but now we must consider

$\displaystyle \frac{1}{n+1}\lim_{y\to{0^+}}y\ln(y)$

$\displaystyle \underbrace{=}_{y=\frac{1}{t}}=\frac{-1}{n+1}\lim_{t\to\infty}\frac{-\ln(t)}{t}$

Which by L'hopital's easily is shown to be zero

$\displaystyle \therefore\quad\int_0^1\ln(y)y^n~dy=\frac{-1}{(n+1)^2}$

So back to our integral we see that we have

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

$\displaystyle =-\sum_{n=1}^{\infty}\frac{1}{n^2}$

$\displaystyle =-\zeta(2)$

$\displaystyle =-\frac{\pi^2}{6}$

$\displaystyle \therefore\quad\boxed{\int_0^1\frac{\ln(1-x)}{x}~dx=\frac{-\pi^2}{6}}$

10. You can thank Arbolis for this post, this one took me a while because of messy algebra. I kept losing "a's" and "b's" and then two steps later looking back and going "Aww!"

$\displaystyle \int_0^{\pi}\frac{dx}{a+b\cos(x)}\quad{a>b>0}$

Now if we let $\displaystyle x=2\arctan\left(t\right)$

Then

$\displaystyle dx=\frac{2}{1+t^2}$

And we know that

$\displaystyle \cos(2\theta)=2\cos^2(\theta)-1$

And we also know that

$\displaystyle \cos\left(\arctan(\theta)\right)=\frac{1}{\sqrt{x^ 2+1}}$

$\displaystyle \Rightarrow\cos\left(2\arctan(t)\right)=\frac{2}{t ^2+1}-1$

So

$\displaystyle a+b\cos\left(2\arctan(t)\right)=a+\frac{2b}{1+t^2}-b$

So we have

$\displaystyle \int\frac{\frac{2dt}{1+t^2}}{a+\frac{2b}{1+t^2}-b}$

$\displaystyle =\int\frac{\frac{2dt}{1+t^2}}{a+\frac{2b}{1+t^2}-b}\cdot\underbrace{\frac{1+t^2}{1+t^2}}_{\text{mul tiplicative identity}}$

$\displaystyle =2\int\frac{dt}{(a-b)t^2+a+b}$

$\displaystyle =\frac{2}{a-b}\int\frac{dt}{t^2+\frac{a+b}{a-b}}$

Now if we let $\displaystyle t=\sqrt{\frac{a+b}{a-b}}\tan(\theta)$

$\displaystyle \Rightarrow{dt=\sqrt{\frac{a+b}{a-b}}\sec^2(\theta)}$

Giving us

$\displaystyle \frac{2}{a-b}\cdot\sqrt{\frac{a-b}{a+b}}\int\frac{sec^2(\theta)}{\tan^2(\theta)+1} ~d\theta$

$\displaystyle =\frac{2}{a-b}\cdot\sqrt{\frac{a-b}{a+b}}\int~d\theta$

$\displaystyle =\frac{2\arctan\left(\sqrt{\frac{a-b}{a+b}}t\right)}{\sqrt{a^2-b^2}}$

Finally backsubbing we get

$\displaystyle \int\frac{dx}{a+b\cos(x)}$

$\displaystyle =\frac{2\arctan\left(\sqrt{\frac{a-b}{a+b}}\tan\left(\frac{x}{2}\right)\right)}{\sqrt {a^2-b^2}}$

So

If we let

$\displaystyle F(x)=\frac{2\arctan\left(\sqrt{\frac{a-b}{a+b}}\tan\left(\frac{x}{2}\right)\right)}{\sqrt {a^2-b^2}}$

Then

$\displaystyle \int_0^{\pi}\frac{dx}{a+b\cos(x)}=F(\pi)-F(0)$

We can obviously see that

$\displaystyle F(0)=0$ since $\displaystyle \tan\left(\frac{0}{2}\right)=0$ making the entire kernel of the arctangent function zero thus making the whole value zero

and we can see that at

$\displaystyle F(\pi)$ we have the kernel as infinity, and the arctangent of infinity is $\displaystyle \frac{\pi}{2}$

So

$\displaystyle F(\pi)=\frac{\pi}{\sqrt{a^2-b^2}}$

$\displaystyle \therefore\quad\boxed{\int_0^{\pi}\frac{dx}{a+b\co s(x)}=\frac{\pi}{\sqrt{a^2-b^2}}}$

And using the same general formula we can show many other integration table favorites

$\displaystyle \int_0^{\frac{\pi}{2}}\frac{dx}{a+b\cos(x)}=\frac{ 2\arctan\left(\sqrt{\frac{a-b}{a+b}}\right)}{\sqrt{a^2-b^2}}$

etc.

11. $\displaystyle \int_0^{\infty}\frac{\arctan(bx)-\arctan(ax)}{x}~dx$

$\displaystyle =\int_0^{\infty}\int_a^{b}\frac{dy}{1+x^2y^2}~dx$

$\displaystyle =\int_a^{b}\int_0^{\infty}\frac{dx}{1+x^2y^2}~dy$

$\displaystyle =\frac{\pi}{2}\int_a^b\frac{1}{y}$

$\displaystyle =\frac{\pi}{2}\ln\left(\frac{b}{a}\right)$

Too tired to explain.

12. Hi!

Int[1/(x^2+e^(-x))]dx from 0 to +oo

(the assignment actually requires to prove if the integral converges on 0,+oo)

thanks!

marine

13. $\displaystyle \int_0^1\ln(x)\ln(1-x)~dx$

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

$\displaystyle =\int_0^1\int_1^y\frac{\ln(x)}{1-y}~dx~dy$

$\displaystyle =\int_0^1\bigg[1+\frac{y\ln(y)}{1-y}\bigg]~dy$

$\displaystyle =\int_0^1\bigg[1+\ln(y)\frac{y-1+1}{1-y}\bigg]~dy$

$\displaystyle =\int_0^1\bigg[1+\ln(y)\left(-1+\frac{1}{1-y}\right)\bigg]~dy$

$\displaystyle \int_0^1\bigg[\frac{\ln(y)}{1-y}-\ln(y)+1\bigg]~dy$

$\displaystyle =\int_0^1\frac{\ln(y)}{1-y}~dy+2$

Now I think I have done this one too, but.

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

$\displaystyle =\int_0^1\sum_{n=0}^{\infty}\ln(y)y^n~dy$

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

$\displaystyle =-\sum_{n=1}^{\infty}\frac{1}{n^2}$

$\displaystyle =-\zeta(2)$

$\displaystyle =\frac{-\pi^2}{6}$

So putting this all together we get

$\displaystyle \boxed{\int_0^1\ln(x)\ln(1-x)~dx=2-\frac{\pi^2}{6}}$

14. With some hints from two people on another website (one of which was an MHF patron ) I am back for a couple more. All integrals that I had hints on I shall put a $\displaystyle \blacksquare$ before. I don't want to steal any ideas!

My hint for this one was that if $\displaystyle a>1$

Then

$\displaystyle \frac{1}{x^a}=\frac{1}{\Gamma(a)}\int_0^{\infty}y^ {a-1}e^{-xy}~dy$

This implies that

$\displaystyle \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\pi}}\int_0^{\in fty}\frac{e^{-yx}}{\sqrt{y}}~dy$

So we may rewrite the following integral as follows

$\displaystyle \int_0^{\infty}\frac{\sin(px)}{\sqrt{x}}~dx$

$\displaystyle =\frac{1}{\sqrt{\pi}}\int_0^{\infty}\int_0^{\infty }\frac{e^{-yx}\sin(px)}{\sqrt{y}}~dy~dx$

$\displaystyle =\frac{\sqrt{\pi}}\int_0^{\infty}\int_0^{\infty}\f rac{e^{-yx}\sin(px)}{\sqrt{y}}~dx~dy$

Now as I showed in the last thread

$\displaystyle \int_0^{\infty}e^{-yx}\sin(px)~dx$

$\displaystyle =\frac{p}{y^2+p^2}$

So we have

$\displaystyle \frac{p}{\sqrt{\pi}}\int_0^{\infty}\frac{dy}{\sqrt {y}(y^2+p^2)}$

Let $\displaystyle z^2=y$

So

$\displaystyle dy=2z~dz$

So we have

$\displaystyle \frac{2p}{\sqrt{\pi}}\int_0^{\infty}\frac{dz}{z^4+ p^2}$

Now this took me an hour just to do, a lot of chances for smudged p's to look like b's etc.

Here you go, remember this one

$\displaystyle \frac{1}{z^4+p^2}=\frac{\sqrt{2}z+2\sqrt{p}}{4p^{\ frac{3}{2}}\left(z^2+\sqrt{2p}z+p\right)}-\frac{\sqrt{2}z-2\sqrt{p}}{4p^{\frac{3}{2}}\left(z^2-\sqrt{2p}z+p\right)}$

Now we must break this up into four terms, two of which will be ln when integrated and two that will be arctan.

Now I will not take the time to type it up but

$\displaystyle \int\frac{dz}{z^4+p^2}$

$\displaystyle =\frac{\sqrt{2}\arctan\left(\frac{\sqrt{2}z-\sqrt{p}}{\sqrt{p}}\right)}{4p^{\frac{3}{2}}}$$\displaystyle +\frac{\arctan\left(\frac{\sqrt{2}z+\sqrt{p}}{\sqr t{p}}\right)}{4p^{\frac{3}{2}}}-\frac{\sqrt{2}\ln\left(z^2-\sqrt{2p}z+p\right)}{8p^{\frac{3}{2}}}+\frac{\sqrt {2}\ln\left(z^2+\sqrt{2p}z+p\right)}{8p^{\frac{3}{ 2}}}=F(z) Now lets consider \displaystyle F(0) now we can see the first two terms drop out since once the zero is put in we will have identical terms except the kernel of the second arctan will be negative. But we know that arctan is odd so this will transfer out giving a term subtracted by an identical term, thus zero. Now for the second terms its similar. Once we drop all the zeros we have two identical terms being subtracted So we can see that \displaystyle F(0)=0 Now let us consider \displaystyle F(\infty) We see that \displaystyle \frac{\sqrt{2}z-2\sqrt{p}}{\sqrt{p}}\sim\frac{\sqrt{2}z+2\sqrt{p}} {\sqrt{p}}\sim\sqrt{2}z\quad\text{As }z\to\infty So \displaystyle \lim_{z\to\infty}\frac{\sqrt{2}\arctan\left(\frac{ \sqrt{2}z+2\sqrt{p}}{\sqrt{p}}\right)}{4p^{\frac{3 }{2}}}\sim$$\displaystyle \lim_{z\to\infty}\frac{\sqrt{2}\arctan\left(\frac{ \sqrt{2}z-2\sqrt{p}}{\sqrt{p}}\right)}{4p^{\frac{3}{2}}}\sim \lim_{z\to\infty}\frac{\sqrt{2}\arctan(\sqrt{2}z)} {4p^{\frac{3}{2}}}=\frac{\sqrt{2}\pi}{8{p}^{\frac{ 3}{2}}}$

And similarly

$\displaystyle \lim_{z\to\infty}\frac{\sqrt{2}\ln(z^2+\sqrt{2p}z+ p)}{8p^{\frac{3}{2}}}\sim\lim_{z\to\infty}\frac{\s qrt{2}\ln(z^2-\sqrt{2p}z+p)}{8p^{\frac{3}{2}}}$

So now consider what asymptotic equivalence means, it means that as z goes to infinity the functions become equivalent.

So if we let $\displaystyle a$ represent the two arctan functions which are equivalent as discussed above and $\displaystyle b$ represent the two natural log functions we have that

$\displaystyle \lim_{z\to\infty}F(z)=a+a-b+b=2a$

And as we have discussed

$\displaystyle a=\frac{\sqrt{2}\pi}{8{p}^{\frac{3}{2}}}$

This implies that

$\displaystyle \lim_{z\to\infty}F(z)=2a=\frac{\sqrt{2}\pi}{4{p}^{ \frac{3}{2}}}$

Now seeing that

$\displaystyle \int_0^{\infty}\frac{dz}{z^4+p^2}=F(\infty)-F(0)$

We see that

$\displaystyle \int_0^{\infty}\frac{dz}{z^4+p^2}$

$\displaystyle =F(\infty)-F(0)$

$\displaystyle =\frac{\sqrt{2}\pi}{4{p}^{\frac{3}{2}}}-0$

$\displaystyle =\frac{\sqrt{2}\pi}{4{p}^{\frac{3}{2}}}$

So we can see that

$\displaystyle \frac{2p}{\sqrt{\pi}}\int_0^{\infty}\frac{dz}{z^4+ p^2}$

$\displaystyle =\frac{\sqrt{2\pi}}{2\sqrt{p}}$

So after much work we arrive at

$\displaystyle \boxed{\int_0^{\infty}\frac{\sin(px)}{\sqrt{x}}~dx =\frac{\sqrt{2\pi}}{2\sqrt{p}}}$

This implies that

$\displaystyle \int_0^{\infty}\frac{\sin(x)}{\sqrt{x}}~dx=\sqrt{\ frac{\pi}{2}}$

15. 'Now we must break this up into four terms, two of which will be ln when integrated and two that will be arctan.

Now I will not take the time to type it up but':

1/(z^4+p^2) = 1/[(z^2+ p - z*sqrt(2p))*(z^2+ p + z*sqrt(2p)], because

z^4 + p^2 = z^4 + 2pz^2 + p^2 - 2pz^2 = (z^2+p)^2 - (z*sqrt(2p))^2 =

= (z^2+p - z*sqrt(2p))*(z^2+p + z*sqrt(2p)

we have fractioned the denominator, now we can do the partial fractions, solve a system and come up with the result of Mathstud28

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