# Complicated Integration

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• Dec 2nd 2007, 02:38 AM
brenorocha
Complicated Integration
Is there anybody here who know hows to solve any of the problems below? I've tried some methods but none of then worked.

$
\int \frac{cos(x)}{x^2+a^2}\,dx
$

$
\int \frac{xsin(x)}{x^2+a^2}\,dx
$

$
\int \frac{x^\frac{1}{3}}{x^2+1}\,dx
$

$
\int \frac{log(x)}{x^2+1}\,dx
$

It would be really helpfull if anybody knows any of the problems above.
Thanks for the help
• Dec 2nd 2007, 07:05 AM
ThePerfectHacker
Maybe you have the limit $-\infty$ to $\infty$ in that case some complex analysis will do it.
• Dec 2nd 2007, 07:14 AM
brenorocha
Ops...
I forgot to mention that, the limits are 0 and $\infty$. Isn't there a way of solving that without putting complex numbers involved?
I mean, I haven't learnt it yet, anyway, could you please post the solution of any of then with that complex analysis
• Dec 2nd 2007, 07:36 AM
Krizalid
Quote:

Originally Posted by brenorocha
$
\int_0^\infty \frac{log(x)}{x^2+1}\,dx
$

This one is fairly known. Substitute $u=\ln x.$

You'll get an odd function on the symmetric interval $(-\infty,\infty),$ so the expected answer is zero.
• Dec 2nd 2007, 07:43 AM
ThePerfectHacker
Quote:

Originally Posted by brenorocha
$
\int \frac{cos(x)}{x^2+a^2}\,dx
$

$\int_{-\infty}^{\infty} \frac{\cos x}{x^2+a^2} dx$ where $a>0$.
Define,
$f(z) = \frac{e^{iz}}{z^2+a^2}$
And integrate over a large semi-circle with radius $R>0$ with $R\to \infty$.
As a result we will have,
$\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+a^2}dx = 2\pi i \left( \mbox{res}(f,ai) \right)$
Thus,
$\int_{-\infty}^{\infty} \frac{\cos x}{x^2+a^2}dx + i \int_{-\infty}^{\infty}\frac{\sin x}{x^2+a^2}dx = 2\pi i \left( \frac{e^{i^2a}}{ai+ai}\right) = 2\pi i \left( \frac{e^{-a}}{2ai} \right)= \frac{\pi}{ae^a}$
• Dec 2nd 2007, 08:39 AM
brenorocha
Thanks
Thank you for the help...
I'll try to solve the others now.
Just to notice, how do I calculate the res() that is in your solution?
The other ( $\frac{1}{3}\int_0^{\infty}\frac{1}{t(t^6+1)} dt$ ), how can I solve it? Trigonometric substitution? Parts? or Partial fraction? which is the easiest?
• Dec 2nd 2007, 09:08 AM
PaulRS
Quote:

Originally Posted by brenorocha
The other ( $\frac{1}{3}\int_0^{\infty}\frac{1}{t(t^6+1)} dt$ ), how can I solve it? Trigonometric substitution? Parts? or Partial fraction? which is the easiest?

I'd let $u=t^6$

But you don't get that integral with the subsitution $x^{1/3}=t$

$\int{\frac{x^{1/3}}{x^2+1}}dx=3\cdot{\int{\frac{t^3}{t^6+1}}}dt$
• Dec 2nd 2007, 10:27 AM
Krizalid
$\varphi=\int_0^\infty {\frac{{x^{1/3} }}
{{x^2 + 1}}\,dx} .$

As said, make the substitution $x=u^3,$

$\varphi = \int_0^\infty {\frac{{x^{1/3} }}
{{x^2 + 1}}\,dx} = \int_0^\infty {\frac{{3u^3 }}
{{u^6 + 1}}\,du} .$

Define another substitution $u=\frac1v,$

$\varphi = \int_0^\infty {\frac{{3v}}
{{v^6 + 1}}\,dv} .$

By symmetry,

$\varphi = \int_0^\infty {\frac{{3x^3 }}
{{x^6 + 1}}\,dx} = \int_0^\infty {\frac{{3x}}
{{x^6 + 1}}\,dx} .$

So, $2\varphi = 3\int_0^\infty {\frac{{x\left( {x^2 + 1} \right)}}
{{\left( {x^2 + 1} \right)\left( {x^4 - x^2 + 1} \right)}}\,dx} = 3\int_0^\infty {\frac{x}
{{x^4 - x^2 + 1}}} .$

Final substitution $\mu=x^2,$

$2\varphi = \frac{3}
{2}\int_0^\infty {\frac{{d\mu }}
{{\mu ^2 - \mu + 1}}} = \frac{3}
{2} \cdot \frac{{4\pi }}
{{3\sqrt 3 }}.$

And we happily get $\varphi = \frac{\pi }
{{\sqrt 3 }}.$

:D:D
• Dec 2nd 2007, 11:34 AM
topsquark
Quote:

Originally Posted by Krizalid
By symmetry,

$\varphi = \int_0^\infty {\frac{{3x^3 }}
{{x^6 + 1}}\,dx} = \int_0^\infty {\frac{{3x}}
{{x^6 + 1}}\,dx} .$

What is the symmetry argument here? (I'm going to kick myself for not spotting the answer, I just know it! (Doh) )

-Dan
• Dec 2nd 2007, 11:54 AM
brenorocha
Quote:

Originally Posted by Krizalid
Define another substitution $u=\frac1v,$

$\varphi = \int_0^\infty {\frac{{3v}}
{{v^6 + 1}}\,dv} .$

Isn't there a signal missing?
$\varphi = \int_0^\infty {\frac{{-3v}}
{{v^6 + 1}}\,dv} .$
• Dec 2nd 2007, 12:24 PM
topsquark
Quote:

Originally Posted by brenorocha
Isn't there a signal missing?
$\varphi = \int_0^\infty {\frac{{-3v}}
{{v^6 + 1}}\,dv} .$

No. The integral after the substitution comes out to be
$\int_{\infty}^0 \frac{ \frac{1}{v} }{ \frac{1}{v^6} + 1 } \cdot -\frac{dv}{v^2}$

The negative sign restores the integral to a $\int_0^{\infty}$ type.

-Dan
• Dec 2nd 2007, 01:18 PM
brenorocha
Quote:

Originally Posted by Krizalid
$\int_0^\infty \frac{log(x)}{x^2+1}\,dx$

Quote:

Originally Posted by Krizalid

This one is fairly known. Substitute $u=\ln x.$

You'll get an odd function on the symmetric interval $(-\infty,\infty),$ so the expected answer is zero.

Making that substitution I got:

$\int_0^\infty \frac{u e^u}{e^{2u}+1}\,du$

But I can't see why it would be 0. Is that really that obvious????
• Dec 2nd 2007, 03:02 PM
Krizalid
Quote:

Originally Posted by topsquark
What is the symmetry argument here?

It's just turnin' back $u,v$ into $x.$ There's no difference between the integrals

$\int_0^\infty {\frac{{3u^3 }}
{{u^6 + 1}}\,du} = \int_0^\infty {\frac{{3x^3 }}
{{x^6 + 1}}\,dx} .$

The value of the integral is always the same.

Quote:

Originally Posted by brenorocha

Making that substitution I got:

$\int_0^\infty \frac{u e^u}{e^{2u}+1}\,du$

Make again the substitution and check the lower integration limit.
• Dec 2nd 2007, 03:21 PM
brenorocha
That's right... I forgot to change the limits... sorry
Thanks for the help
there is just one left now...

$\int \frac{xsin(x)}{x^2+a^2}\,dx$
• Dec 2nd 2007, 04:33 PM
topsquark
Quote:

Originally Posted by Krizalid
It's just turnin' back $u,v$ into $x.$ There's no difference between the integrals

$\int_0^\infty {\frac{{3u^3 }}
{{u^6 + 1}}\,du} = \int_0^\infty {\frac{{3x^3 }}
{{x^6 + 1}}\,dx} .$

The value of the integral is always the same.

Like I said: (Doh)

Thanks.

-Dan
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