1. Originally Posted by Mathstud28
You mean $\displaystyle \frac{\pi}{2e}$ and yes....but I sense some Complex Analysis in the mix here

EDIT: You corrected yourself.
your integral = $\displaystyle 2\int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx - 2\int_0^{\infty} \frac{x \sin x}{(x^2 + 1)^2} \ dx,$ and by parts in the second integral gives us: your integral = $\displaystyle \int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx=\int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx$ why?

and i guess i don't have to prove this very well-known fact that $\displaystyle \int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx = \frac{\pi}{2e}.$

2. I may as well finish how I done it. Overkill, yes. I wish I would have saw that Frullani thing right off. I am familiar with it.It did not occur to me. I will remember to look around for it from now on.

By letting $\displaystyle f(x)=\frac{e^{x}}{e^{x}+1}+C$

I left $\displaystyle f(ax)-f(bx)=\frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}dx$

Then setting $\displaystyle I(u)=\int_{0}^{u}\frac{f(ax)}{x}dx-\int_{0}^{u}\frac{f(bx)}{x}dx$

This is OK if $\displaystyle \lim_{v\to 0}\frac{f(v)}{v}$ exists. Which works

if C = -1/2.

With C = -1/2, we get:

$\displaystyle I(u)=\int_{0}^{au}\frac{f(v)}{v}dv-\int_{0}^{bu}\frac{f(v)}{v}dv$

$\displaystyle =\int_{au}^{bu}\frac{f(v)}{v}dv$

To save typng, I went on and eventually wound up with said solution.

I like all of the methods. That is why I like to post an occasional integral.

3. Originally Posted by NonCommAlg
your integral = $\displaystyle \int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx=\int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx$ why?
I assume the "why" was meant to be answere by me?

Answer: Since $\displaystyle \int_0^{\infty}\frac{x\sin(x)}{x^2+1}$ is even we have that $\displaystyle \int_0^{\infty}\frac{x\sin(x)}{x^2+1}=\frac{1}{2}\ int_{-\infty}^{\infty}\frac{x\sin(x)}{x^2+1}$

Integrating by parts letting $\displaystyle dv=\sin(x)$ and $\displaystyle u=\frac{x}{x^2+1}$ gives

$\displaystyle \frac{1}{2}\int_{-\infty}^{\infty}\frac{x\sin(x)}{x^2+1}=$$\displaystyle \frac{1}{2}\left\{-\cos(x)\left(\frac{2}{x^2+1}-\frac{1}{x^2+1}\right)\right\}\bigg|_{x=-\infty}^{x=\infty}-\frac{1}{2}\int_{-\infty}^{\infty}\left\{\frac{-2\cos(x)}{(x^2+1)^2}+\frac{\cos(x)}{x^2+1}\right\} dx$

Now $\displaystyle \int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)^2}=\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx$ Why?

So we get

\displaystyle \begin{aligned}\frac{-1}{2}\int_{-\infty}^{\infty}\left\{\frac{-2\cos(x)}{(x^2+1)^2}+\frac{\cos(x)}{x^2+1}\right\} &=\int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)^2}-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}\\ &=\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx\\ &=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}\\ &=\int_0^{\infty}\frac{\cos(x)}{x^2+1}dx\end{align ed}

4. As for the well-known fact, see

http://www.mathhelpforum.com/math-he...tegration.html

last post.

5. Originally Posted by Krizalid
As for the well-known fact, see

http://www.mathhelpforum.com/math-he...tegration.html

last post.
Ahh that is one of my favorite ways to solve it...it is obviously much easier to use CA though...Ill be sure to use it when I learn it

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