3

Question

The Feynman's integral trick (a.k.a. differentiation under the integral) proves to be a powerful tool when dealing with definite integrals. However, I have been unable to find a similar extension of Feynman's trick in the case of indefinite integrals. Thus, I conclude that such an extension is not probably possible, which begs the question why? Also, is there such a similar (not the same) trick for inefinite integrals?

My reasoning

I do have a plausible reason why this trick doesn't work with indefinite integrals. According to my understanding, we are unable to determine the arbitrary constant of integration (which could even probably be a function of the integration variable) when re-integrating the differentiated integral. Though, it would be better if anyone could either verify, or falsify, this, or provide a better explanation.

What motivated me to ask the above question?

The following paragraphs talk about the particular integral which resulted into the above line of thought. Unrelated to the crux of the question, but might be helpful for some context.
Do note that I am not asking for a solution which evaluates the following integral, because neither do I not know how to evaluate it, nor is the following integral the spotlight of the question. The question is a general one, concerning every indefinite integral which could possibly solved by Feynman's trick.

I encountered the following indefinite integral

$$I=\int \frac{\arctan x}{x^4}\mathrm dx$$

Now I do know of a method to evaluate this integral using integration by parts, and it's quite straightforward, however, this time my first thought was to introduce a parameter $b$ such that

$$I(b)=\int \frac{\arctan (bx)}{x^4}\mathrm dx$$

Now differentiating $I(b)$ with respect to $b$ yields

$$\frac{\mathrm d [I(b)]}{\mathrm db}=\int \frac{\mathrm dx}{(1+b^2x^2)x^3}$$

Now substituting $1/x^2$ as $t$, we can easily solve the resultant integral, finally yielding

$$\frac{\mathrm d [I(b)]}{\mathrm db}=-\frac{1}{2x^2}+\frac{b^2}{2}\ln\left(\frac{1}{x^2}+b^2\right)+c'$$

And now I am pretty much clueless about what to do next. And this is where I got curious about such a technique for indefinite integrals.

FWIW, the value of the integral comes out to be (obtained using integration by parts) :

$$I=-\frac{\arctan x}{3x^2}-\frac{1}{6} \ln\left(1+\frac{1}{x^2}\right)-\frac{1}{6x^2} + c$$

  • 1
    why not just transform your indefinite integral into a definite one $I(b,x)=\int_{x_0}^x f(bt)\mathop{dt}$ temporarily ? – zwim Jul 31 '20 at 17:18
  • @zwim Hmm... Let's say I do it. What next? I can't see any way to re-integrate the final equation even with this assumption. –  Jul 31 '20 at 17:19
  • 1
    like integrating $\ln(x)$ you start with $x\ln(x)$ then $x^2\ln(x)$, for $b^2\ln(b^2)$ start with $b^3\ln(b^2)$ and add terms (missing after derivation). – zwim Jul 31 '20 at 17:34
  • @zwim Alright. But then what about the integration constant. How do I determine that? –  Jul 31 '20 at 17:35
  • 1
    Please stick to either $a$ or $b$ or it will be confusing. Also you should have $I'(b)=-\frac{1}{2x^2}+\frac{b^2}{2}\ln\left(\frac{1}{x^2}+b^2\right)+\color{red}{C_1}$ there. Now just integral back as $I(b)=\int I'(b) db$ (which will be something like ($f(b,x)+C_1b+C_2$) and set $b=1$. Also note that a sum of two constants is just another constant, so $C_1+C_2=C$.// Unfortunately for this particular case I don't think it's useful to use Feynman's trick. – Zacky Jul 31 '20 at 17:56
  • 2
    sorry for putting a link, I should leave. https://docdro.id/Y6tcvbX you see that applying Feynman trick (i.e. $g(1)$) and calculating directly ($f(1)$) lead to the same result. The constant $C$ is all the stuff containing $x0$. And you can notice the rest is the same as indefinite integral $I$ you wrote above. – zwim Jul 31 '20 at 17:57
  • @ThreeSidedCoin Oops! My bad, corrected :-) –  Jul 31 '20 at 18:03
  • Very relevant conversation, essentially answering the question: https://chat.stackexchange.com/rooms/36/conversation/feynmans-trick-for-indefinite-integrals –  Jul 31 '20 at 18:15
  • Still, I think it's better to stick to what zwim suggested above, I rarely encountered indefinite integrals where this is useful, but this integral is definitely one case (originally it was this). – Zacky Jul 31 '20 at 19:02

0 Answers0