A quick integral. That's the video.

[@or4cl363]

How did you develop such a genuine interest for math

[@JPiMaths]

@or4cl363 that's a great question, I have no idea... I think I just enjoyed solving problems!

[@kostasch5686]

Excellent. You can achieve the same result by splitting the initial integral from 0 to 1 and 1 to infinity then substitute one of them with 1/u.

[@JPiMaths]

ah nice! That's sometimes a really useful trick!

[@bhodson7954]

One question. When you switch u for x at 1m32. Surely the two seperate integrals now have a different version of x? I appreciate u is a dummy variable but it is defined as u=1/x (x=1/u). Sorry if that is a dumb question! I've just done a whole set of different substitutions and checked the evaluation on Wolfram alpha. No doubt it works, but why! 😂

[@badamson]

it really doesn’t matter how you label it. We could have switched the variable from u -> t at the end and it still would have been the same, or not switched it at all, and kept it at u. Yes, u = 1/x, but we are integrating with respect to u now, not x. We account for the problem you’re raising when we transform dx->du already. You could imagine two unrelated problems with two integrals that have identical form, but the variable is a different symbol. Since the answer is just a number, and is not in terms of x or u, clearly these are identical. Remember, the variable only exists within the scope of that integral. Like, there’s no hidden details in the definite integral, if the two integrals look exactly the same, they are exactly the same. You could imagine working out one manually, and working out the other one manually and clearly we would have no difference in the result.

[@JPiMaths]

@@bhodson7954 good question. It's the same reason that the integral from 0 to 2 of x^3 dx is the same as the integral from 0 to 2 of u^3 du. When you sub in the limits, it doesn't matter what the letters you use are.

[@bhodson7954]

@@JPiMaths Thanks for helping. As an extension, I plotted both 1/(1+x^4) and x^2/(1+x^4) Obviously they are both different. I think the key here in me understanding my original question is that what the process shows is that the *integrals* within the range 0 to infinity are the same, *not* the individual two original algebraic fractions. The algebraic manipulation you carried out follows rules. I also evaluated the integral between 1 and 4 of 1/(1+x^4) and after making the same u=1/x substitution, this was exactly the same as the integral between 0.25 and 1 of x^2/(1+x^4). Thanks again for a great youtube channel

[@raghvendrasingh1289]

👍 also if we take positive sign in place of negative then ans = π√2/2 similarly integral of (x^8) (1 - x^6)/(1+x)^14 in [0, infinity] = 0

[@JPiMaths]

@@raghvendrasingh1289 interesting... 🤔

[@adiramrakhani]

Cool problem! How did you come up with that substitution?

[@JPiMaths]

@@adiramrakhani Thanks! It's a relatively natural substitution to attempt as you know it will 'flip' the limits of the integral