# Thread: Differentiating through the integral sign

1. ## Differentiating through the integral sign

Here's the question:

By integrating through the integral sign, evaluate $\int_{0}^\infty (x^2 + a)^{-n} dx$ where n is a positive integer and a > 1.

So the standard idea is to let F(a) = $\int_{0}^\infty (x^2 + a)^{-n} dx$ then use (after checking various conditions hold) F'(a) = $\int_{0}^\infty (-n)(x^2 + a)^{-n-1} dx$ by just differentiating through the integral sign, evaluating the integral and solving the resulting differential equation for F(a).

But here we cannot easily work out the integral, so what can we do?

2. Originally Posted by Amanda1990
Here's the question:

By integrating through the integral sign, evaluate $\int_{0}^\infty (x^2 + a)^{-n} dx$ where n is a positive integer and a > 1.

So the standard idea is to let F(a) = $\int_{0}^\infty (x^2 + a)^{-n} dx$ then use (after checking various conditions hold) F'(a) = $\int_{0}^\infty (-n)(x^2 + a)^{-n-1} dx$ by just differentiating through the integral sign, evaluating the integral and solving the resulting differential equation for F(a).

But here we cannot easily work out the integral, so what can we do?
You should try an induction argument: if you let $F_n(a)=\int_0^\infty \frac{dx}{(x^2+a)^n}$, then you proved that $F_n'(a)=-nF_{n+1}(a)$.

Notice that $F_1(a)$ can be evaluated. Then you get $F_2(a)$ by differentiating $F_1(a)$, and $F_3(a)$ by differentiating $F_2(a)$, and so on...

3. Nice idea! I can see exactly how the induction would work but I haven't been able to find a pattern and claim what the result should be in the first place...What would the inductive hypothesis be?

4. Originally Posted by Amanda1990
Nice idea! I can see exactly how the induction would work but I haven't been able to find a pattern and claim what the result should be in the first place...What would the inductive hypothesis be?
Did you get $F_1(a)=\frac{\pi}{2}\frac{1}{\sqrt{a}}$? Then $F_1'(a)=\frac{\pi}{2}\frac{-\frac{1}{2}}{a^{\frac{1}{2}+1}}$, so $F_2(a)= \frac{1}{1}\frac{\pi}{2}\frac{\frac{1}{2}}{a^{\fra c{1}{2}+1}}$. Then $F_3(a)=\frac{1}{2}F_2'(a)=\frac{1}{1\cdot 2}\frac{\pi}{2}\frac{\frac{1}{2}\left(\frac{1}{2}+ 1\right)}{a^{\frac{1}{2}+2}}$.

From there, you can infer $F_n(a)=\frac{1}{1\cdot 2\cdots (n-1)}\frac{\pi}{2}\frac{\frac{1}{2}\left(\frac{1}{2} +1\right)\cdots\left(\frac{1}{2}+(n-2)\right)}{a^{\frac{1}{2}+(n-1)}}$. This will be straightforward to prove using the induction formula. Then you can simplify the expression, like $F_n(a)=\frac{1\cdot 3\cdot 5\cdots(2n-3)}{2^n (n-1)!}\pi=\frac{(2n-3)!}{2^{2n-1}(n-1)!^2}\pi$.