Improper Integral: Solve for \pi\log(x+1)

[DeadOriginal]

Homework Statement

$\int_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy$

Homework Equations

The answer is $\pi\log(x+1)$.

The Attempt at a Solution

I have attempted many different substitutions like $y=\tan\theta$. I have also tried breaking up the log but nothing definitive comes out. Any help would be appreciated. I would prefer to not use any complex analysis.

 

[Mark44]

Homework Statement

$\int_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}$

Homework Equations

The answer is $\pi\log(x+1)$.

The Attempt at a Solution

I have attempted many different substitutions like $y=\tan\theta$. I have also tried breaking up the log but nothing definitive comes out. Any help would be appreciated. I would prefer to not use any complex analysis.

Which is the variable of integration? Without either dx or dy, I can't tell.

 

[DeadOriginal]

Which is the variable of integration? Without either dx or dy, I can't tell.

Sorry. It is now fixed. We are integrating with respect to y.

 

[DeadOriginal]

I am beginning to think that there is a way to show that the integral is equal to $\pi\log(x+1)$ without directly computing the integral. Any thoughts?

 

[SammyS]

I am beginning to think that there is a way to show that the integral is equal to $\pi\log(x+1)$ without directly computing the integral. Any thoughts?

Perhaps, ...

consider that $\displaystyle \ \frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}\ .$

 

[STEMucator]

Perhaps, ...

consider that $\displaystyle \ \frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}\ .$

Yes, this.

Expressing it as an expansion can also work.

 

[DeadOriginal]

Hmm. That looks very interesting. Thanks guys. I will play with that.

 

[DeadOriginal]

I am confused. How do I use the fact that $\frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}$? Shouldn't we be considering the changes in $y$ and not $x$?

 

[STEMucator]

I am confused. How do I use the fact that $\frac{d}{dx}\pi\log(x+1)=\frac{\pi}{x+1}$? Shouldn't we be considering the changes in $y$ and not $x$?

Are you familiar with the expansion for $\frac{1}{1-x}$?

Use that to find an expansion for $\frac{\pi}{1-(-x)}$

What do you get?

 

[DeadOriginal]

I got $\pi\sum\limits_{n=0}^{\infty}(-1)^{n}x^{n}$ but I don't see how I can apply this to solving the integral.

 

[haruspex]

Let the original integral be F(x). What does dF/dx look like? Can you solve the integral wrt y in that?

 

[DeadOriginal]

$F(x)=\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy$
so
$\frac{d}{dx}F(x)=\frac{d}{dx}\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\int\limits_{0}^{\infty}\frac{2xy^{2}}{(y^{2}+1)(x^{2}y^{2}+1)}dy=2x\int\limits_{0}^{\infty}\frac{y^{2}}{(y^{2}+1)(x^{2}y^{2}+1)}dy=2x\int\limits_{0}^{\infty}\left[\frac{Ay+B}{y^{2}+1}+\frac{Cy+D}{x^{2}y^{2}+1}\right]$
where $A,C=0$ and $B=-\frac{1}{1-x^{2}},D=\frac{1}{1-x^{2}}$. Then
$2x\int\limits_{0}^{\infty}\left[\frac{Ay+B}{y^{2}+1}+\frac{Cy+D}{x^{2}y^{2}+1}\right]=\frac{2x}{1-x^{2}}\int\limits_{0}^{\infty}\frac{1}{x^{2}y^{2}+1}dy-\frac{2x}{1-x^{2}}\int\limits_{0}^{\infty}\frac{1}{y^{2}+1}dy=\frac{2x}{1-x^{2}}\left(\frac{\pi}{2x}\right)-\frac{2x}{1-x^{2}}\left(\frac{\pi}{2}\right)=\frac{\pi-x\pi}{1-x^{2}}=\pi\left(\frac{1-x}{1-x^{2}}\right)=\frac{\pi}{x+1}$.

This shows that
$\frac{d}{dx}F(x)=\frac{d}{dx}\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\frac{d}{dx}\pi\log(x+1)$
but I don't see how it shows that
$\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}dy=\pi\log(x+1)$.
When we integrate them both in terms of $x$, would we not have a constant on both sides that could be different?

 

[haruspex]

When we integrate them both in terms of $x$, would we not have a constant on both sides that could be different?

Yes, so the final step is to show that the constant is zero. You have f(x) = g(x) + c for all x. How might you determine c? Any ideas?

 

[DeadOriginal]

Hmm... Let me sit on this for a while. I will post again once I have an idea! Thanks!

 

[DeadOriginal]

Ok. Here's an idea. Took longer to think of than it should of..
Since
$F(x)=\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}=\pi\log(x+1)+C$
we can let $x=0$ such that
$F(0)=\int\limits_{0}^{\infty}\frac{\log(1)}{y^{2}+1}=\pi\log(1)+C=0$.
Hence it must be that $C=0$.

 

[haruspex]

Ok. Here's an idea. Took longer to think of than it should of..
Since
$F(x)=\int\limits_{0}^{\infty}\frac{\log(x^{2}y^{2}+1)}{y^{2}+1}=\pi\log(x+1)+C$
we can let $x=0$ such that
$F(0)=\int\limits_{0}^{\infty}\frac{\log(1)}{y^{2}+1}=\pi\log(1)+C=0$.
Hence it must be that $C=0$.

That'll do it.

 

[DeadOriginal]

Thanks a ton!