Ytukyg

How to solve the following integral:

$$int_{-infty}^{+infty}{frac{1}{x^2-a}dx},$$where $a$ is a constant.

Hint:

You can get rid of the constant with a scaling of the variable, $x=sqrt{|a|}t$ and

$$int_{-infty}^{+infty}{frac{1}{x^2-a}dx}=frac1{sqrt{|a|}}int_{-infty}^{+infty}{frac{1}{t^2-text{sgn }a}dt}.$$

In case $a<0$, you recognize the derivative of the arc tangent.

In case $age0$, you recognize the derivative of the hyperbolic arc tangent or of the inverse function. As these have singularities, the integral can only be computed in the improper sense.

I assume that $a$ is real.

Hints:

1. If $a>0$, then $$frac 1 {x^2-a} = 2sqrt aleft(frac 1 {x-sqrt a} - frac 1 {x+sqrt a}right)$$
   Then try to apply $int frac 1 x mathrm d x = log x + C$.




2. If $a=0$, use $$int x^k mathrm d x = frac 1 {k+1} x^{k+1} + C$$ for $kne -1$.




3. If $a<0$, try $$int frac 1 {x^2+1} mathrm d x = arctan x +C,$$

where $C$ is a constant of integration.

Basically the same as the other answers but more detail:

Case 1 ($a=0$) This one is pretty obvious so yeah

Case 2 ($a>0$) Since $a>0$, the polynomial $x^2-a$ has two real roots. And as you are probably aware, it can be factored: $x^2-a=(x-sqrt{a})(x+sqrt{a})$. So we can use partial fractions.

Since the fraction can be factored, we say that it can be expanded:
$$frac1{(x-sqrt{a})(x+sqrt{a})}=frac{A}{x-sqrt{a}}+frac{B}{x+sqrt{a}}$$
Thus
$$frac1{(x-sqrt{a})(x+sqrt{a})}=frac{(x+sqrt{a})A+(x-sqrt{a})B}{(x-sqrt{a})(x+sqrt{a})}$$
$$1=(x+sqrt{a})A+(x-sqrt{a})B$$
Since this property should hold for all $x$, then it should hold for $x=sqrt{a}$. So, we plug that in:
$$1=(sqrt{a}+sqrt{a})A+(sqrt{a}-sqrt{a})B$$
$$1=2sqrt{a},A$$
$$A=frac1{2sqrt{a}}$$
That was so cool let's do it again but this time with $x=-sqrt{a}$:
$$1=(-sqrt{a}+sqrt{a})A+(-sqrt{a}-sqrt{a})B$$
$$...$$
$$B=-frac1{2sqrt{a}}$$
So we have a fraction expansion:
$$frac1{(x-sqrt{a})(x+sqrt{a})}=frac1{2sqrt{a}}bigg(frac{1}{x-sqrt{a}}-frac{1}{x+sqrt{a}}bigg)$$
For the sake of a detailed answer, I'll show you the integration steps.
$$
begin{align}
intfrac{mathrm dx}{x^2-a}=&frac1{2sqrt{a}}intfrac{mathrm dx}{x-sqrt{a}}-frac1{2sqrt{a}}intfrac{mathrm dx}{x+sqrt{a}}\
=& frac1{2sqrt{a}}log|x-sqrt{a}|-frac1{2sqrt{a}}log|x+sqrt{a}|\
=& frac1{2sqrt{a}}logbigg|frac{x-sqrt{a}}{x+sqrt{a}}bigg|+C\
end{align}
$$
Just remember that $log$ is the natural logarithm (not the base $10$ logarithm)-I write $log$ instead of $ln$ because it looks cooler.

Also you may note that for $a>0$, the definite integral diverges.

Case 3 ($a<0$) We could do this multiple ways. We could do it the way we did in Case 2, but then we would be dealing with complex numbers, and while that's fun, it's not as fun as a trig sub. Trig subs are seriously great check this out:

$$I=intfrac{mathrm dx}{x^2-a}$$
For this we do something that isn't obvious at first. We preform the substitution
$x=sqrt{a}sec(u)$. Thus $mathrm dx=sqrt{a},sec(u)tan(u) mathrm du$. Hence
$$
begin{align}
I=&intfrac{sqrt{a},sec(u)tan(u)}{(sqrt{a},sec(u))^2-a}mathrm du\
=&intfrac{sqrt{a},sec(u)tan(u)}{asec^2(u)-a}mathrm du\
=&frac{sqrt{a}}{a}intfrac{sec(u)tan(u)}{sec^2(u)-1}mathrm du\
=&frac1{sqrt{a}}intfrac{sec(u)tan(u)}{tan^2(u)}mathrm du\
=&frac1{sqrt{a}}intfrac{sec(u)}{tan(u)}mathrm du\
=&frac1{sqrt{a}}intfrac{mathrm du}{sin(u)}\
=&-frac1{sqrt{a}}log|cot(u)+csc(u)|\
I=&-frac1{sqrt{a}}logbigg|frac{x+sqrt{a}}{sqrt{x^2-a}}bigg|+C\
end{align}
$$

And then just use the fundamental theorem of calculus to find the definite integral.

I'm sorry if you already knew these integration techniques, I just like to do really detailed answers just to make sure that everyone gets the intuitive nature behind the solution.