1. Originally Posted by Mathstud28
Nice Paul!

Could someone give me a starting hint on the two following ones?

$\int_0^{\infty}\frac{\sin^2(x)}{x^2}~dx$
Is the answer for this $\frac{\pi}2$?

2. Yes.

3. Originally Posted by Krizalid
Express $\sin^2x$ into another form by using $\cos2x.$ Consider that $\frac{1}{x^{2}}=\int_{0}^{\infty }{ye^{-xy}\,dy}.$

You recently found a parameter for the integrand.
Dangit, that should have been obvious thanks Krizalid.

$\int\frac{\sin^2(px)}{x^2}~dx$

So

$\frac{1}{2}\int\frac{1-\cos(2px)}{x^2}~dx$

Now let $u=2px\Rightarrow\frac{du}{2p}=dx$

$\frac{1}{4p}\int\frac{1-\cos(u)}{\left(\frac{u}{2p}\right)^2}~du=p\int_0^{ \infty}\frac{1-\cos(u)}{u^2}=$

$=p\int_0^{\infty}\int_0^{\infty}(1-\cos(u))\cdot{ye^{-yu}}~dy~du$

$=p\int_0^{\infty}\int_0^{\infty}(1-\cos(u))\cdot{ye^{-yu}}~du~dy$

$=p\int_0^{\infty}\bigg[\left(\frac{\cos(u)y^2}{y^2+1}-\frac{\sin(u)y}{y^2+1}-1\right)e^{-xy}\bigg]\bigg|_0^{\infty}~dy$

$=p\int_0^{\infty}\frac{dy}{y^2+1}$

$=p\bigg[\arctan(\infty)-\arctan(0)\bigg]=\frac{\pi{p}}{2}$

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

4. $\int_0^{\infty}\frac{\cos(ax)-\cos(bx)}{x}~dx$

$=\int_0^{\infty}\int_a^{b}\sin(yx)~dy~dx$

$=\int_a^{b}\int_0^{\infty}\sin(yx)~dx~dy$

$=\int_a^{b}\frac{-\cos(xy)}{y}\bigg|_0^{\infty}~dy$

$\text{Here is when I am lost}~=\int_a^{b}\frac{dy}{y}$

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

$\Rightarrow{\sin(\infty)=1}$

$\Rightarrow{\pi|\infty}$??

5. Oops, I misread the problem. If the denominator was $xe^x,$ the integral does converge.

6. Originally Posted by Mathstud28
I thought this was kind of coincidental, this ones not on the list so I hope I am correct.

$\int_0^{\infty}\frac{\cos(ax)-\cos(bx)}{e^{px}x}~dx$

$=\int_0^{\infty}e^{-px}\int_a^{b}\sin(yx)~dy~dx$

$=\int_0^{\infty}\int_a^{b}e^{-px}\sin(yx)~dy~dx$

$=\int_a^{b}\int_0^{\infty}e^{-px}\sin(yx)~dx~dy$

$=\int_a^{b}\bigg[\frac{-e^{-px}\cos(xy)y}{y^2+p^2}-\frac{pe^{-px}\sin(xy)}{y^2+p^2}\bigg]\bigg|_0^{\infty}~dy$

$=\int_a^{b}\frac{y}{y^2+p^2}~dy$

$=\frac{1}{2}\bigg[\ln(b^2+p^2)-\ln(a^2+p^2)\bigg]$

$=\boxed{\ln\left(\sqrt{\frac{b^2+p^2}{a^2+p^2}}\ri ght)}$

$\Rightarrow\int_0^{\infty}\cos(px)\frac{e^{-ax}-e^{-bx}}{x}~dx=\int_0^{\infty}e^{-px}\frac{\cos(ax)-\cos(bx)}{x}~dx$

That blows my mind, I hope I am right, otherwise this will be dissapointing.
Originally Posted by Krizalid
Oops, I misread the problem. If the denominator was $xe^x,$ the integral does converge.
Yeah I did that integral already ^^

But this one website says that the answer is $\ln\left(\frac{b}{a}\right)$. Anyone know why?

7. I'm sorry, I have my head in another place.

You can't reverse integration order. (Check it why.) If you consider $\frac1x=\int_0^\infty e^{-xy}\,dy,$ that'd give ya the expected answer.

8. Originally Posted by Isomorphism
Is the answer for this $\frac{\pi}2$?
Originally Posted by Krizalid
Yes.
Ok I have an alternate way, other than double integrals. The moment I see my sinc functions, my signal processing harmones pester me to use fourier transforms.

Here is the fourier transform way.

Note: $\text{sinc}(x) = \frac{\sin (\pi x)}{\pi x}$

From fourier transform theory, $F(\text{sinc}(t)) = \text{rect}(f)$, where rect denotes the function defined by:

$\text{rect}(f) = \begin{array}{cc} 1 & -\frac12 \leq f \leq \frac12 \\ 0 & \text{elsewhere} \end{array}$

By the convolution-multiplication theorem, multiplication in one domain gives convlution in another domain.

Thus $F(\text{sinc}^2(x)) = F(\text{sinc}(x) \times \text{sinc}(x)) = \text{rect}(f) * \text{rect}(f)$

Convolving the rect function is easy so i will skip it. It gives the triangle function:

$\text{tri}(f) = \begin{array}{cc} 1 + f & -1 \leq f \leq 0 \\ 1 - f & 0 \leq f \leq 1 \\0 & \text{elsewhere} \end{array}$

Now by definition of fourier transform:

$\int_{-\infty}^{\infty} \text{sinc}^2(t) e^{-2\pi i f t} \, dt = \text{tri}(f)$

To compute the integral we want, put f=0.

$\int_{-\infty}^{\infty} \text{sinc}^2(t)\, dt = \text{tri}(0) = 1$
$\int_{-\infty}^{\infty} \left(\frac{\sin (\pi t)}{\pi t}\right)^2 \, dt = 1$
$2\int_{0}^{\infty} \left(\frac{\sin (\pi t)}{\pi t}\right)^2 \, dt = 1$

Now substitute $x = \pi t$, to get:

$\frac{2}{\pi}\int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^2 \, dx = 1$

$\int_{0}^{\infty}\frac{\sin^2 x}{x^2} \, dx = \frac{\pi}2$

9. Originally Posted by Krizalid
I'm sorry, I have my head in another place.

You can't reverse integration order. (Check it why.) If you consider $\frac1x=\int_0^\infty e^{-xy}\,dy,$ that'd give ya the expected answer.
Thats a lot of work

$\int_0^{\infty}\frac{\cos(ax)-cos(bx)}{x}~dx$

= $=\int_0^{\infty}\int_0^{\infty}e^{-yx}\left(\cos(ax)-\cos(bx)\right)~dy~dx$

$=\int_0^{\infty}\int_0^{\infty}e^{-yx}\left(\cos(ax)-\cos(bx)\right)~dx~dy$

$=\int_0^{\infty}\bigg[\left(\frac{-\cos(ax)y}{y^2+a^2}+\frac{a\sin(ax)}{y^2+a^2}+\fra c{\cos(bx)y}{y^2+b^2}-\frac{b\sin(xy)}{y^2+b^2}\right)\bigg]\bigg|_0^{\infty}~dy$

$=\int_0^{\infty}\bigg[\frac{y}{y^2+b^2}-\frac{y}{y^2+a^2}\bigg]~dy$

$=\ln\left(\sqrt{\frac{y^2+b^2}{y^2+a^2}}\right)\bi gg|_0^{\infty}$

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

10. I have tried and failed to solve this integral... Can anybody give me a hint?

$\int_0^{\pi} \frac{\sin^2 (nx)}{\sin^2 x}\, dx$ where n is a positive integer.

11. Originally Posted by Isomorphism
I have tried and failed to solve this integral... Can anybody give me a hint?

$\int_0^{\pi} \frac{\sin^2 (nx)}{\sin^2 x}\, dx$ where n is a positive integer.
Hint one

trig identity

Hint two

trig sub

12. Originally Posted by Isomorphism
I have tried and failed to solve this integral... Can anybody give me a hint?

$\int_0^{\pi} \frac{\sin^2 (nx)}{\sin^2 x}\, dx$ where n is a positive integer.
It appeared in this thread if you want to check. By the way it may also be done by (changes to white) using Euler's Identity and a simple algebraic identity

13. Hint 3
induction ?
i mean, relation between rank n and n+1..

14. Here are a couple more I don't have the answers to, so I would appreciate if someone with an advanced knowledge of integrals...or a very nice computer system ...could verify them?

$\int_0^{\infty}\frac{e^{-ax}\sin(px)}{x}~dx\quad\boxed{1}$

Now we know that

$\frac{1}{x}=\int_0^{\infty}e^{-yx}~dy$

So we may rewrite $\boxed{1}$ as follows

$\int_0^{\infty}\frac{e^{-ax}\sin(px)}{x}~dx$

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

$=\int_0^{\infty}\int_0^{\infty}e^{-x(y+a)}\sin(px)~dy~dx$

Now since the region of integration is rectangular by Fubini's Theorem we may rewrite $\boxed{2}$ as follows

$\int_0^{\infty}\int_0^{\infty}e^{-x(y+a)}\sin(px)~dx~dy$

Now by two iterations of Integration by Parts (just for a note to those who wish to duplicate this, for simplicities sake call $y+a=z$) the inner iterated integral is equivalent to

$=\int_0^{\infty}\bigg[\left(\frac{-p\cos(px)}{(y+a)^2+p^2}-\frac{\sin(px)(y+a)}{(y+a)^2+p^2}\right)\bigg]\bigg|_0^{\infty}\quad\boxed{3}$

Now seeing that as $x\to\infty\Rightarrow\boxed{3}\to{0}$ due to the overpowering effect of $e^{-x(y+a)}$, and evaluating at zero we get

$=\int_0^{\infty}\frac{p}{(y+a)^2+p^2}~dy$

Now for the calculation of

$\int\frac{p}{(y+a)^2+p^2}~dy$

The substitution of

$z=y+a\Rightarrow{dz=dy}$

Gives

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

Now letting $z=p\tan(\theta)$

Gives us the final answer of

$\arctan\left(\frac{y+a}{p}\right)$

So we then have that

$\int_0^{\infty}\frac{p}{(y+a)^2+p^2}~dy$

$=\arctan\left(\frac{y+a}{p}\right)\bigg|_0^{\infty }$

$=\arctan\left(\infty\right)-\arctan\left(\frac{0+a}{p}\right)$

$=\frac{\pi}{2}-\arctan\left(\frac{a}{p}\right)$

$={\rm{arcot}}\left(\frac{a}{p}\right)$

$\therefore\quad\boxed{\int_0^{\infty}\frac{e^{-ax}\sin(px)}{x}~dx={\rm{arcot}}\left(\frac{a}{p}\r ight)}$

15. $\int_0^{\infty}\frac{\cos(ax)-\cos(bx)}{x^2}~dx$

$=\int_0^{\infty}\int_0^{\infty}ye^{-yx}\left(\cos(ax)-\cos(bx)\right)~dy~dx$

$=\int_0^{\infty}\int_0^{\infty}ye^{-yx}\left(\cos(ax)-\cos(bx)\right)~dx~dy$

$=\int_0^{\infty}\bigg[\left(\frac{-\cos(ax)y^2}{y^2+a^2}+\frac{a\sin(ax)y}{y^2+a^2}+\ frac{\cos(bx)y^2}{y^2+b^2}-\frac{by\sin(xy)}{y^2+b^2}\right)e^{-xy}\bigg]\bigg|_0^{\infty}~dy$

$=\int_0^{\infty}\bigg[\frac{a^2}{y^2+a^2}-\frac{b^2}{y^2+b^2}\bigg]~dy$

$=\bigg[a\arctan\left(\frac{y}{a}\right)-b\arctan\left(\frac{y}{b}\right)\bigg|_0^{\infty}$

$=\frac{\pi(a-b)}{2}\quad\blacksquare$

Page 2 of 6 First 123456 Last