$$ I = \int_{0}^{\infty} \frac{1}{ (1+ax^2)^{m+1}} dx$$ Then, $$ I'(a) = -(m+1) \int_{0}^{\infty} \frac{2ax}{(1+ax^2)^{2m+2} } dx$$

so that

$$I'(a) = \frac{ m+1}{2(2m-1)} [ (1+ax^2)^{1-2m}]_{0}^{\infty}$$

Now what do I do? I am finding it difficult to proceed

  • $\begingroup$ Are you missing something in your second line? Did you forget to finish up an integration by parts reasoning? $\endgroup$ Commented Jun 11, 2020 at 13:50
  • $\begingroup$ I'll add more stuff $\endgroup$ Commented Jun 11, 2020 at 13:56
  • $\begingroup$ I added one more step pls check $\endgroup$ Commented Jun 11, 2020 at 13:57
  • $\begingroup$ I see, you took the derivative wrong. $x$ is supposed to be a constant. $\endgroup$ Commented Jun 11, 2020 at 13:59

3 Answers 3


$I'(a)$ should really be

$$I'(a) = -(m+1)\int_0^\infty \frac{x^2}{(1+ax^2)^{m+2}}\:dx$$

Then use integration by parts:

$$I'(a) = \frac{x}{2a(1+ax^2)^{m+1}}\Bigr|_0^\infty - \frac{1}{2a}\int_0^\infty \frac{1}{(1+ax^2)^{m+1}}\:dx$$

which means that

$$2aI' + I = 0$$

Can you take it from here?

I'll still leave the general solution to you. However, one thing you'll immediately find is that the usual candidates for initial values don't tell us anything new as $I(0) \to \infty$ and $I(\infty) \to \infty$. Instead we'll try to find $I(1)$:

$$I(1) = \int_0^\infty \frac{1}{(1+x^2)^{m+1}}\:dx$$

The trick is to let $x = \tan \theta \implies dx = \sec^2 \theta \:d\theta$

$$I(1) = \int_0^\frac{\pi}{2} \cos^{2m}\theta\:d\theta$$

Since the power is even, we can use symmetry to say that

$$\int_0^\frac{\pi}{2} \cos^{2m}\theta\:d\theta = \frac{1}{4}\int_0^{2\pi} \cos^{2m}\theta\:d\theta$$

Then use Euler's formula and the binomial expansion to get that

$$ = \frac{1}{4^{m+1}}\sum_{k=0}^{2m}{2m \choose k} \int_0^{2\pi} e^{i2(m-k)\theta}\:d\theta$$

All of the integrals will evaluate to $0$ except when $k=m$, leaving us with the only surviving term being

$$I(1)=\frac{2\pi}{4^{m+1}}{2m \choose m}$$

  • $\begingroup$ I get $ I = \frac{C'}{2a}$ , im not sure where to evaluaute it now $\endgroup$ Commented Jun 11, 2020 at 14:04
  • $\begingroup$ @DDD4C4U That...is not right. You should really review your ODE solving techniques. Remember that $a$ is not a constant, it is the variable. $\endgroup$ Commented Jun 11, 2020 at 14:05
  • $\begingroup$ I don't know why I tried few times and I am not getting the soln $\endgroup$ Commented Jun 11, 2020 at 14:11
  • $\begingroup$ ok is it $ \frac{C}{ a^{\frac{1}{2} }$ $\endgroup$ Commented Jun 11, 2020 at 14:12
  • 1
    $\begingroup$ @NinadMunshi ;-) Pleased to hear you liked the post. Stay safe and healthy my friend! $\endgroup$
    – Mark Viola
    Commented Jun 11, 2020 at 15:04

The integral of interest $$\displaystyle I(a;m)=\int_0^\infty \frac1{(1+ax^2)^{m+1}}\,dx$$ converges for $a>0$ and $m>-1/2$.

We now present an approach that is valid for non-integer values of $m>-1/2$.

Feynman's Trick is actually a waste of effort here. All we need to do is enforce the substitution $\sqrt{a} x\mapsto x$ to find that

$$I(a;m)=\frac1{\sqrt a}\int_0^\infty \frac1{(1+x^2)^{m+1}}\,dx$$ Then, we can value the integral

$$C(m)=\int_0^\infty \frac1{(1+x^2)^{m+1}}\,dx\tag1$$

by enforcing a second substitution $x^2\mapsto x$ as shown subsequently.

If we wish to use Feyman's trick, we follow the solution posted by @ninadmunshi, to arrive at $$\frac{\partial }{\partial a}I(a;m)+\frac1{2a}I(a;m)=0\tag2$$

The general solution to $(2)$ is $I(a;m)=C(m)a^{-1/2}$, which is the solution we found already by making the simple substitution $\sqrt a x\mapsto x$.

To find $C(m)$ we seek to evaluate $I(1;m)$. Proceeding, we enforce the substitution $x^2\mapsto x$ to obtain for $m>-1/2$

$$\begin{align} C(m)&=I(1;m)\\\\ &=\int_0^\infty \frac{1}{(1+x^2)^{m+1}}\,dx\\\\ &=\frac12\int_0^\infty \frac{1}{x^{1/2}(1+x)^{m+1}}\,dx\\\\ &=\frac12 B\left(1/2,m+1/2\right)\\\\ &= \frac{\sqrt\pi\,\Gamma(m+1/2)}{2\Gamma(m+1)} \end{align}$$

Therefore, we find

$$\bbox[5px,border:2px solid #C0A000]{I(a;m)=\frac{\sqrt\pi\,\Gamma(m+1/2)}{2\sqrt{a}\,\Gamma(m+1)}}$$

NOTE: Solution for integer valued of $m$

If $m\in \mathbb{N}$, then $\Gamma(m+1)=m!$ and $\Gamma(m+1/2)=\frac{2^{1-2m}\sqrt{\pi}(2m-1)!}{(m-1)!}$ and

$$\bbox[5px,border:2px solid #C0A000]{I(a;m)=\frac{\pi (2m-1)!}{4^m m!(m-1)!\sqrt{a}}}$$

  • $\begingroup$ @DDD4c4u Please let me know how I can improve my answer. I really want to give you the best answer I can. Note that Feynman's Trick really buys nothing here. $\endgroup$
    – Mark Viola
    Commented Aug 18, 2020 at 2:23

So we have ${I(a) = \int_{0}^{\infty}\frac{1}{(1+ax^2)^{m+1}}dx}$. The derivative is ${-(m+1)\int_{0}^{\infty}\frac{x^2}{(1+ax^2)^{m+2}}}$ (remember that $x$ is a constant with respect to $a$). Now, from integration by parts (with ${dv=\frac{-(m+1)x}{(1+ax^2)^{m+2}}, u=x}$) we get $${\left(\frac{x}{2a(1+ax^2)^{m+1}}\right)_{0}^{\infty} - \frac{1}{2a}\int_{0}^{\infty}\frac{1}{(1+ax^2)^{m+1}}dx}$$

Obviously the limit will be zero, and hence we have the relation ${I'(a) = \frac{-1}{2a}I(a)}$. You can see this is a very simple separable differential equation:

$${\int\frac{dI}{I}=\frac{-1}{2}\int\frac{1}{a}da}$$ Hence we have ${\ln(I)=-0.5\ln(a)+C}$, and so


The only thing left to do is find the constant ${A}$. This is not really trivial to do, and in general will be in terms of $m$. Take ${I(1)}$:


Let's make the substitution ${x = \tan(t)}$. This gives


Since we are dealing with even powers of cosine, we can write


A formula exists for this, see: Integral of $\int_0^{2\pi}\cos^n(x)\,dx$. :


And so finally,


Giving our final answer


Edit: ah I see many people already beat me to it, I'm slow at writing LaTeX haha. The way I like to think of Feynmans trick (which is just a special case of the Leibniz rule for integration) is that we essentially are parameterising our integral, and looking at how our integral changes with respect to our new variable. If we can then "pin down" our integral to a value at a certain point (that is much easier for us to evaluate), we can deduce what value the integral will take on for different values. Of course if we take multiple derivatives with Feynmans trick, the more constants you will have to account for...

  • $\begingroup$ @MarkViola, should I remove my answer? $\endgroup$ Commented Jun 11, 2020 at 14:55
  • $\begingroup$ No, don't remove it. You have added some commentary that is useful. $\endgroup$
    – Mark Viola
    Commented Jun 11, 2020 at 15:00
  • $\begingroup$ @MarkViola Okaydokey, I'll leave it here :D $\endgroup$ Commented Jun 11, 2020 at 15:05

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