1. ## Integrals

Hey mathstud, you know what might be fun?. Go to the link I posted above and actually derive the solutions they give for the integrals. You have probably done that already.

Some of them are probably very challenging.
You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}$

Ok my first move was to do the obvious double integral

$\displaystyle \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy$

So we have

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

Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

$\displaystyle \iint_{R}e^{-yx}\csc(px)~dA$

Where $\displaystyle R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b} }\right\}$

From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

So

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

Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

parts-no deal

Letting $\displaystyle J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx$-no go

Any suggestions?

I also tried some crazy thing where I let

$\displaystyle J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx$

And found that the integral I want is a constant difference from $\displaystyle J'(0)$

2. How are you getting from

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}$

to

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

Shouldn't it be

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

No suggestions on how to proceed with this, but $\displaystyle \frac{1}{\csc x} = \sin x$

3. Originally Posted by Mathstud28
You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}$

Ok my first move was to do the obvious double integral

$\displaystyle \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy$

So we have

$\displaystyle \color{red}\int_0^{\infty}\csc(px)\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}\csc(px)~dy~dx$

Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

$\displaystyle \iint_{R}e^{-yx}\csc(px)~dA$

Where $\displaystyle R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b} }\right\}$

From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

So

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

Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

parts-no deal

Letting $\displaystyle J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx$-no go

Any suggestions?

I also tried some crazy thing where I let

$\displaystyle J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx$

And found that the integral I want is a constant difference from $\displaystyle J'(0)$
I believe that this is where you're screwing up...

I think that it should be $\displaystyle \int_0^{\infty}{\color{red}\sin(px)}\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}{\color{red}\sin(px)}~dy~dx$

because : $\displaystyle \frac{e^{-ab}-e^{-bx}}{x\csc(px)}=\frac{e^{-ab}-e^{-bx}}{x}\cdot\frac{1}{\csc(px)}=\frac{(e^{-ab}-e^{-bx}){\color{red}\sin(px)}}{x}$

EDIT : icemanfan beat me to it...

4. Originally Posted by Chris L T521
I believe that this is where you're screwing up...

I think that it should be $\displaystyle \int_0^{\infty}{\color{red}\sin(px)}\int_a^{b}e^{-yx}~dy~dx=\int_0^{\infty}\int_a^{b}e^{-yx}{\color{red}\sin(px)}~dy~dx$

because : $\displaystyle \frac{e^{-ab}-e^{-bx}}{x\csc(px)}=\frac{e^{-ab}-e^{-bx}}{x}\cdot\frac{1}{\csc(px)}=\frac{(e^{-ab}-e^{-bx}){\color{red}\sin(px)}}{x}$

EDIT : icemanfan beat me to it...
<-- with an f. Now I feel stupid for copying it down wrong

Well

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

By Integration by Parts twice then adding to both sides.

So now as x goes to infinity obviously the $\displaystyle e^{-yx}$ overpowers and it goes to zero, and when $\displaystyle x=0$ the above equals

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

So we have that

$\displaystyle \int_0^{\infty}e^{-yx}\sin(px)~dx=\frac{p}{y^2+p^2}$

So now we must perform

$\displaystyle \int_a^{b}\frac{p}{x^2+p^2}~dx$

Rewriting this as

$\displaystyle \frac{1}{p}\int\frac{dx}{\left(\frac{x}{p}\right)^ 2+1}~dx$

And letting $\displaystyle u=\frac{x}{p}\Rightarrow{p\cdot{du}=dx}$

So we have

$\displaystyle \int\frac{du}{u^2+1}=\arctan(u)$

Back subbing we get

$\displaystyle \arctan\left(\frac{x}{p}\right)$

$\displaystyle \therefore\quad\int_a^{b}\frac{p}{x^2+p^2}~dx=\arc tan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)$

$\displaystyle \therefore\quad\boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}~dx=\arctan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)}$

Seeing that I did a-b opposed to b-a this is what they had

5. Originally Posted by Mathstud28
You know me well sir. I already was there deriving them, now most of them I can see what to do. But one is alluding me . Now I haven't given it much time but

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}$

Ok my first move was to do the obvious double integral

$\displaystyle \frac{e^{-ax}-e^{-bx}}{x}=\int_a^{b}e^{-yx}dy$

So we have

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

Now rewriting this as (this is so people who aren't super accustomed to double integrals can see what I am doing)

$\displaystyle \iint_{R}e^{-yx}\csc(px)~dA$

Where $\displaystyle R=\left\{(x,y):0\leq{x}\leq\infty~{a\leq{y}\leq{b} }\right\}$

From here it is obvious that this region is a rectangle, so Fubini's Theorem applies

So

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

Now this inner integral is bothering me, I cannot seem to get it in the eight minutes I have tried

parts-no deal

Letting $\displaystyle J(\theta)=\int_0^{\infty}e^{-yx}\csc(px)e^{-\theta\sin(px)}dx$-no go

Any suggestions?

I also tried some crazy thing where I let

$\displaystyle J(\theta)=\int_0^{\infty}\frac{e^{-xy\cos(px\theta)}}{\sin(px)}~dx$

And found that the integral I want is a constant difference from $\displaystyle J'(0)$
My 89 doesn't give a solution and Mathematica tells me this:

Why don't you try to solve this using infinite series?

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{k\csc(px)}\,dx=\sum_{k=0}^{\infty}\left[e^{-ak}-e^{-bk}\right]\cdot\sum_{k=0}^{\infty}\left[\frac{\sin(k)}{k}\right]$

You know the power series expansions better than I do. Experiment with this and let me know how you do. I'll try to remember what I can to see I can evaluate the integral...

Wait...can this be related to the zeta function, by any chance [I'm probably wrong...just throwing ideas out there]?

--Chris

6. Originally Posted by Mathstud28
<-- with an f. Now I feel stupid for copying it down wrong

Well

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

By Integration by Parts twice then adding to both sides.

So now as x goes to infinity obviously the $\displaystyle e^{-yx}$ overpowers and it goes to zero, and when $\displaystyle x=0$ the above equals

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

So we have that

$\displaystyle \int_0^{\infty}e^{-yx}\sin(px)~dx=\frac{p}{y^2+p^2}$

So now we must perform

$\displaystyle \int_a^{b}\frac{p}{x^2+p^2}~dx$

Rewriting this as

$\displaystyle \frac{1}{p}\int\frac{dx}{\left(\frac{x}{p}\right)^ 2+1}~dx$

And letting $\displaystyle u=\frac{x}{p}\Rightarrow{p\cdot{du}=dx}$

So we have

$\displaystyle \int\frac{du}{u^2+1}=\arctan(u)$

Back subbing we get

$\displaystyle \arctan\left(\frac{x}{p}\right)$

$\displaystyle \therefore\quad\int_a^{b}\frac{p}{x^2+p^2}~dx=\arc tan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)$

$\displaystyle \therefore\quad\boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\csc(px)}~dx=\arctan\left(\frac{b}{p}\right)-\arctan\left(\frac{a}{p}\right)}$

Seeing that I did a-b opposed to b-a this is what they had
Originally Posted by Chris L T521
My 89 doesn't give a solution and Mathematica tells me this:

Why don't you try to solve this using infinite series?

$\displaystyle \int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{k\csc(px)}\,dx=\sum_{k=0}^{\infty}\left[e^{-ak}-e^{-bk}\right]\cdot\sum_{k=0}^{\infty}\left[\frac{\sin(k)}{k}\right]$

You know the power series expansions better than I do. Experiment with this and let me know how you do. I'll try to remember what I can to see I can evaluate the integral...

Wait...can this be related to the zeta function, by any chance [I'm probably wrong...just throwing ideas out there]?

--Chris
Um...ok...disregard my last post...

These are interesting integrals!

--Chris

7. Originally Posted by Chris L T521
Um...ok...disregard my last post...

These are interesting integrals!

--Chris
Don't feel bad, you may not have seen that, but I made up a new identity

$\displaystyle \frac{1}{\csc(x)}=\csc(x)$

8. The other one

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

Ok once again seeing that

$\displaystyle \frac{e^{-ax}-e^{-bx}}{x}=\int_{a}^{b}e^{-yx}dy$

We have that $\displaystyle \boxed{1}$ is equivalent to

$\displaystyle \boxed{2}\quad\int_0^{\infty}\int_a^{b}e^{-yx}\cos(px)~dy~dx$

Once again seeing the region of integration is a rectangle we may switch the integration order without changing the limits by Fubini's Theorem

$\displaystyle \boxed{2}=\int_a^{b}\int_0^{\infty}e^{-yx}\cos(px)~dx~dy$

Now once again by double integration by parts and algebra

$\displaystyle \int{e^{-yx}\cos(px)}dx=e^{-yx}\bigg[\frac{p\sin(px)}{y^2+p^2}-\frac{\cos(px)y}{y^2+p^2}\bigg]\quad\boxed{3}$

Now $\displaystyle \lim_{x\to\infty}\boxed{3}=0$

and $\displaystyle \boxed{3}$ evaluated at zero is $\displaystyle \frac{-y}{y^2+p^2}$

so $\displaystyle \boxed{4}\quad\int_0^{\infty}e^{-yx}\cos(px)~dx=\frac{y}{y^2+p^2}$

So using $\displaystyle \boxed{4}$

We see that

$\displaystyle \int_a^b\int_0^{\infty}e^{-yx}\cos(px)~dx~dy=\int_a^{b}\frac{y}{y^2+p^2}~dy$

Now for $\displaystyle \int\frac{y}{y^2+p^2}~dy$

If we let $\displaystyle u=y^2+p$ then $\displaystyle du=2y\Rightarrow\frac{du}{2}=y$

So we have

$\displaystyle \frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln(u)$

So back subbing we get

$\displaystyle \frac{1}{2}\ln\left(y^2+p^2\right)$

So

$\displaystyle \int_a^{b}\frac{y}{y^2+p^2}=\frac{1}{2}\bigg[\ln(b^2+p^2)-\ln(a^2+p^2)\bigg]=\frac{1}{2}\ln\left(\frac{b^2+p^2}{a^2+p^2}\right )=\ln\left(\sqrt{\frac{b^2+p^2}{a^2+p^2}}\right)\q uad\boxed{5}$

Now putting $\displaystyle \boxed{1},\boxed{2},\boxed{3},\boxed{4},\boxed{5}$ together we get

$\displaystyle \boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\sec(px)}~dx=\ln\left(\sqrt{\frac{b^2+p^2}{a ^2+p^2}}\right)}\quad\blacksquare$

9. Mmm... but express the integrand into complex form, the rest follows. (This is worth than applyin' partial integration twice.)

10. $\displaystyle \int_0^{1}\frac{\ln(1+x)}{x}~dx\quad$

So seeing that since we can rewrite this as
$\displaystyle \int_0^{1}\frac{dx}{x}\cdot\sum_{n=0}^{\infty}\fra c{(-1)^nx^{n+1}}{n+1}$

$\displaystyle =\int_0^{1}\sum_{n=0}^{\infty}\frac{(-1)x^{n}}{n+1}$

$\displaystyle =\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{(n+1)^2}\bigg|_0^{1}$

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

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

Now we know that

$\displaystyle \left(1-2^{1-x}\right)\zeta(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}$

So we can see that

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\left(1-2^{1-2}\right)\zeta(2)$

$\displaystyle =\frac{1}{2}\cdot\frac{\pi^2}{6}$

$\displaystyle =\frac{\pi^2}{12}$

$\displaystyle \therefore\quad\boxed{\int_0^{1}\frac{\ln(1+x)}{x} ~dx=\frac{\pi^2}{12}}$

I feel as though I have done this before...

11. I want to do this one integral, but first I need to do another one

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

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

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

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

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

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

$\displaystyle =-\left(1-2^{1-2}\right)\cdot\frac{\pi^2}{6}=\frac{-\pi^2}{12}$

$\displaystyle \therefore\quad\int_0^1\frac{\ln(x)}{x+1}~dx=\frac {-\pi^2}{12}\quad\boxed{1}$

So now that that ancillary integral is out of the way I will move onto the integral of interest

$\displaystyle \int_0^{1}\ln(x)\ln(1+x)~dx$

First we see that

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

So we have that

$\displaystyle \int_0^{1}\ln(x)\ln(1+x)~dx=\int_0^{1}\int_0^{x}\f rac{\ln(x)}{1+y}~dy~dx$

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

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

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

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

$\displaystyle =\int_0^{1}\frac{\ln(y)}{y+1}~dy+\bigg[-y\ln(y)+y-2\ln|y+1|+y\bigg]\bigg|_0^{1}$

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

Now using $\displaystyle \boxed{1}$

We can finally conclude that

$\displaystyle \boxed{\int_0^{1}\ln(x)\ln(1+x)~dx=2-2\ln(2)-\frac{\pi^2}{12}}$

12. Originally Posted by Mathstud28
$\displaystyle \boxed{\int_0^{1}\ln(x)\ln(1+x)~dx=2-2\ln(2)-\frac{\pi^2}{12}}$
It may also be proved as follows

$\displaystyle \sum\limits_{k = 1}^{n} {\tfrac{{\left( { - 1} \right)^{k + 1} }} {k} \cdot x^k } + R_n \left( x \right) = \ln \left( {x + 1} \right)$ where $\displaystyle \left( { - 1} \right)^n \cdot \int_0^x {\tfrac{{t^n }} {{1 + t}}} dt = R_n \left( x \right)$ thus, if x>0, we have: $\displaystyle \left| {R_n \left( x \right)} \right| \leqslant \tfrac{{x^n }} {n} = \int_0^x {\tfrac{{t^n }} {t}} dt$

$\displaystyle \sum\limits_{k = 1}^{n} {\tfrac{{\left( { - 1} \right)^{k + 1} }} {k} \cdot x^k \cdot{\ln(x)} } +\ln(x)\cdot R_n \left( x \right) = \ln(x)\cdot \ln \left( {x + 1} \right)$ where $\displaystyle \left| {\ln \left( x \right) \cdot R_n \left( x \right)} \right| \leqslant - \tfrac{{x^n }} {n} \cdot \ln \left( x \right)$ if $\displaystyle x\in(0, 1]$

Integrate, recalling the linearity of the integral and $\displaystyle \int_0^1 {x^k \cdot \ln \left( x \right)dx} = - \tfrac{1} {{\left( {k + 1} \right)^2 }}$

$\displaystyle \sum\limits_{k = 1}^n {\tfrac{{\left( { - 1} \right)^k }} {{k \cdot \left( {k + 1} \right)^2 }}} + \int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} = \int_0^1 {\ln \left( x \right) \cdot \ln \left( {x + 1} \right)dx}$ (1)

$\displaystyle \left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \leqslant \int_0^1 {\left| {\ln \left( x \right) \cdot R_n \left( x \right)} \right|dx} \leqslant - \tfrac{1} {n}\int_0^1 {x^n \cdot \ln \left( x \right)dx}$ thus: $\displaystyle \left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \leqslant \tfrac{1} {{n \cdot \left( {n + 1} \right)^2 }}$

Thus $\displaystyle \left| {\int_0^1 {\ln \left( x \right) \cdot R_n \left( x \right)dx} } \right| \to 0$ as n tends to infinity and then $\displaystyle \sum\limits_{k = 1}^\infty {\tfrac{{\left( { - 1} \right)^k }} {{k \cdot \left( {k + 1} \right)^2 }}} = \int_0^1 {\ln \left( x \right) \cdot \ln \left( {x + 1} \right)dx}$

Now note that: $\displaystyle \tfrac{{\left( { - 1} \right)^k }} {{k \cdot \left( {k + 1} \right)^2 }} = \tfrac{{\left( { - 1} \right)^k }} {k} + \tfrac{{\left( { - 1} \right)^{k + 1} }} {{k + 1}} - \tfrac{{\left( { - 1} \right)^k }} {{\left( {k + 1} \right)^2 }}$ thus: $\displaystyle \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }} {{k \cdot \left( {k + 1} \right)^2 }}} = \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }} {k}} + \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^{k + 1} }} {{k + 1}}} - \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }} {{\left( {k + 1} \right)^2 }}}$ (summed in that order according to (1))

Now since : $\displaystyle \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }} {k}} = - \ln \left( 2 \right);\sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^{k + 1} }} {{k + 1}}} = - \ln \left( 2 \right) + 1$ and $\displaystyle \sum\limits_{k \geqslant 1} {\tfrac{{\left( { - 1} \right)^k }} {{\left( {k + 1} \right)^2 }}} = 1 - \tfrac{{\pi ^2 }} {{12}}$ we get the result

13. Nice Paul!

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

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

Still Lebniz's rule?

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

I would say double integration but I dont know the cosine of infinity

14. Originally Posted by Mathstud28
$\displaystyle \boxed{\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x\sec(px)}~dx=\ln\left(\sqrt{\frac{b^2+p^2}{a ^2+p^2}}\right)}\quad\blacksquare$
I thought this was kind of coincidental, this ones not on the list so I hope I am correct.

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

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

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

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

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

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

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

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

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

That blows my mind, I hope I am right, otherwise this will be dissapointing.

15. Originally Posted by Mathstud28

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

$\displaystyle \int_0^{\infty}\frac{\sin^2(x)}{x^2}~dx$
Express $\displaystyle \sin^2x$ into another form by using $\displaystyle \cos2x.$ Consider that $\displaystyle \frac{1}{x^{2}}=\int_{0}^{\infty }{ye^{-xy}\,dy}.$

Originally Posted by Mathstud28

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

I would say double integration but I dont know the cosine of infinity
You recently found a parameter for the integrand.

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