How can I solve this integral involving x, y, and s?

[muzialis]

Hi All,

I am having some troble with the following integral

$$\int_{-c}^{c} \frac{x^3}{(x^2+(y-s)^2)^2}\mathrm{d}s$$

Many thanks as usual

 

[Simon Bridge]

What's the problem?

Please show your best attempt, annotated so we can see your thought process.

 

[arildno]

To give you a hint:
For some constant "a", functions f and g, do you know of functions that satisfies the identity:
$$a^{2}+f^{2}=g^{2}$$

 

[muzialis]

Arildno,

thanks for your hint. I suggest functions $$f = \sqrt{log x}$$ and $$g = \sqrt{log (x/e^{a^2})}$$.

Simon,

My best attempt so far:
$$\int \frac{1}{(x^2 + (y-s)^2)^2} \mathrm{d}s$$, change variable y-s = z,
$$\int \frac{-1}{(x^2 + z^2)*(x^2+z^2)} \mathrm{d}s =$$
and now I admit buckling under the pressure of the looming partial fraction attempt, trying to get to integrate terms such as $$\int \frac{1}{A+x^2}\mathrm{d}x$$...

Many thanks

 

[Simon Bridge]

I take it that the x and y have nothing to do with the s then?
They are basically just constants for this integration?

Arildno,
thanks for your hint. I suggest functions $$f = \sqrt{\ln x}$$ and $$g = \sqrt{\ln (x/e^{a^2})}$$.

Because: $$a^2+f^2=a^2 + \ln |x| = \ln|e^{a^2}|+\ln|x| = \ln|xe^{a^2}| \neq g^2$$
(did you make a typo?) OK - so, having found a pair of functions that have the suggested property - try using one of them as a substitution in the integrand...

$$\int \frac{1}{(x^2 + (y-s)^2)^2} \mathrm{d}s$$, change variable y-s = z,
$$\int \frac{-1}{(x^2 + z^2)(x^2+z^2)} \mathrm{d}s =$$

... so did you try your substitution?

Note that $x^2+z^2 = (x+z)(x-z)$
... which will help with the partial fractions.

Personally I'd start with the very first line and use a substitution.

I believe you have misunderstood the hint ...

How about trig functions:
$\sin^2\theta+\cos^2\theta=1$ and $1+\tan^2\theta = \sec^2\theta$

The idea is to use one of them to make a substitution.
i.e. starting from $a^2-x^2$ if you substitute $x=a\sin\theta$ then: $$a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2\theta$$

... it's probably easier than the partial fractions.

 

[Mark44]

My first thought was a trig substitution. You can make things simpler by bringing the x³ term outside the integral.
$$x^3\int_{-c}^{c} \frac{ds}{(x^2+(y-s)^2)^2}$$

Note that (y - s)² = (s - y)², so you can make a slight change like so:
$$x^3\int_{-c}^{c} \frac{ds}{(x^2+(s - y)^2)^2}$$

Then by letting z = s - y, dz = ds, you have:
$$x^3\int_{-c}^{c} \frac{dz}{(x^2+ z^2)^2}$$

then a trig substitution, and off you go...

 

[Simon Bridge]

Mark44: I think you misplaced a minus sign.
[edit] oh I see you exploited that (-x)^2=x^2

I figured trig substitutions too ... so did arildno.

You can do the trig sub right at the start with s=y-x<trig fn> without having to do the initial z=s-y step.
Although it is easier to evaluate - the back-substitution gets messy - needing a table of trig identities for things like sin(arctan(x)).

Partial fraction approach is initially more labor intensive but looks like it gets a more useable result right away.
I guess it depends what you're used to.

 

[Mark44]

Mark44: I think you misplaced a minus sign.
[edit] oh I see you exploited that (-x)^2=x^2

I figured trig substitutions too ... so did arildno.

It wasn't obvious that that was where arildno was going.

You can do the trig sub right at the start with s=y-x<trig fn> without having to do the initial z=s-y step.

s = y - x? Must be a typo.

Although it is easier to evaluate - the back-substitution gets messy - needing a table of trig identities for things like sin(arctan(x)).

You don't need a table if you can draw a right triangle.

Let θ = arctan(x). Label a right triangle with the side opposite θ of length x, and the adjacent side of length 1. Then the hypotenuse is √(x² + 1). It's straightforward to get sin(θ); i.e., sin(arctan(x)).

Partial fraction approach is initially more labor intensive but looks like it gets a more useable result right away.
I guess it depends what you're used to.

There's more than one way to skin a cat. However, when I see a sum of squares, I immediately begin to salivate "trig substitution."

 

[Simon Bridge]

It wasn't obvious that that was where arildno was going.
s = y - x? Must be a typo.

x multipied by an appropriate trig function.
I didn't want to spell it out any more than I had to.

If T(θ) is a trig function - then a substitution of form: s=y-xT(θ)
That way (y-s)=xT(θ) and ds = -xT'(θ) ... which, admittedly gives you a minus sign to carry around.

You don't need a table if you can draw a right triangle.

Good point.

There's more than one way to skin a cat. However, when I see a sum of squares, I immediately begin to salivate "trig substitution."

I know - me too.

Now to hear from OP.

 

[arildno]

Actually, I'm rather a hyperbolic guy myself, but trigs work as well.

 

[Simon Bridge]

Actually, I'm rather a hyperbolic guy myself, but trigs work as well.

That's OK - you can get medicine for that :)

 

[muzialis]

Many thanks to all of you for help, very useful and appreciated.
Simon Bridge, yes my solution for the question posed by Arildno was wrong, not a typo but considering g as the constant, again my error.
Thanks again