Although I’m a lawyer, this is a math account. Integrals and number theory. I do math, not politics. Let’s do more math. I love music and baseball. Roses too 🌹
30 Days of Feynman—Day 1
For 30 days, I plan to publish integral solutions using the Feynman trick of differentiating under the integral sign.
Please feel free to reply with other solution methods, or perhaps more efficient ways of writing solutions using the Feynman trick.
This interesting integral begins with the golden ratio (phi) and ends at 1. Note that I used a hyperbolic trig substitution, instead of the standard one.
Here’s my solution to an integral posted here by Prof Strogatz (
@stevenstrogatz
) about a year and a half ago. I found it when I was rummaging through X for something totally unrelated. I’ll put the original post in the first reply if you’re interested in some background info.
Another YouTube channel that I’m recommending is Owls Math. He solved the following problem using the King property and trig identities. My trig-free solution involves an interesting property of definite integrals, as illustrated in my reply.
30 Days of Feynman—Day 2
For 30 days, I plan to publish integral solutions using the Feynman trick of differentiating under the integral sign.
Please feel free to reply with other solution methods, or perhaps more efficient ways of writing solutions using the Feynman trick.
These are integral formulas I derived a week or two ago, using Laplace transforms. It’s actually much easier to derive these formulas using the gamma function, Euler’s reflection formula, and Euler’s most famous formula. You basically get 2 integral formulas for the price of one.
Here’s another twofer, which often results when you evaluate integrals using contour integration. Integral J was recently published under the account of
@Diarytells
, and solved by
@robolsmath
, using another method.
30 Days of Feynman—Day 19
All three integral formulas presented below may be derived using only the results from Day 18. In the first two integrals, I regard a as the parameter. I use a well-known trig identity to set up the third integral.
30 Days of Feynman—Day 10
For 30 days, I’m publishing solutions involving differentiating under the integral sign.
As we go, I am including fewer details, to keep the solutions a bit shorter. For example, here I left the relatively simple PFD exercise on scratch paper. 📝
How to derive 2 and 3 angle sum/difference identities using Euler’s formula and de Moivre’s theorem. This is a continuation from the exercise I started yesterday.
30 Days of Feynman—Day 12
Solutions involving differentiating under the integral sign. I’m including more details in this solution.
In a reply, I evaluated the intermediate integral using the tangent full-angle substitution, a close relative of the Weierstrass substitution.
30 Days of Feynman—Day 6
This is the integral that PK posted earlier this morning on his YouTube channel. I recommend that you watch it because he used an entirely different technique.
As a reminder, for 30 days, I plan to publish integral solutions using the Feynman trick.
30 Days of Feynman—Day 3
For 30 days, I plan to publish integral solutions using the Feynman trick of differentiating under the integral sign.
Please feel free to reply with other solution methods, or perhaps more efficient ways of writing solutions using the Feynman trick.
30 Days of Feynman—Day 4
For 30 days, I plan to publish integral solutions using the Feynman trick of differentiating under the integral sign.
Please feel free to reply with other solution methods, or perhaps more efficient ways of writing solutions using the Feynman trick.
My solution to the integral I posted two days ago. It assumes basic knowledge of analytic number theory. I recently posted links to Wikipedia pages that provide this knowledge. The symbol like a cursive N is the Dirichlet eta function. The lowercase gamma is the Euler constant.
30 Days of Feynman—Day 9
For 30 days, I’m publishing solutions involving differentiating under the integral sign.
As we go, I am including fewer details, to keep the solutions a bit shorter. For example, here I left the relatively simple PFD exercise on scratch paper. 📝
I took an integral recently solved by
@robolsmath
, but I used an Euler substitution. I got a much different-looking, though still correct, anti-derivative. Euler substitutions generally look ugly at first, but you quickly get some nice cancellations.
@iampatriottom
@maggieNYT
They got him loaded up with steroids. So, he's going to say and do a lot of stupid shit between now and Saturday. Save this tweet.
30 Days of Feynman—Day 5
2 quick ones today.
As a reminder, for 30 days, I plan to publish integral solutions using the Feynman trick. Please feel free to reply with other solution methods, or perhaps more efficient ways of writing solutions using the Feynman trick.
30 Days of Feynman—Day 7
I am republishing as a reply to this integral the product-to-sum trig identity that I derived using Euler’s formula.
As a reminder, for 30 days—though not necessarily consecutive—I plan to publish integral solutions using the Feynman trick.
40 Days of Feynman—Day 25
This integral formula uses the result from Day 24. It also includes integral J, which is toward the bottom of the page. These formulas may also be derived by contour integration.
Никола́й Ива́нович Лобаче́вский (Lobachevski), better known as the founder of a non-Euclidean geometry, found the time to derive an integral formula that makes quick work of integrals like the one shown below:
Using Euler’s formula to derive two related integral formulas. This exercise is related to the exercises from last week. The idea, once again, is to equate the real part with the real part, and the imaginary with the imaginary.
30 Days of Feynman—Day 8
The golden ratio appears in the solution. In the first integral, I used the tangent full-angle substitution. In the last integral, I used a u sub (u=root(1+alpha)).
For 30 days, I’m publishing solutions involving differentiating under the integral sign.
Integrals K & L are follow-ups to the integrals I posted yesterday. I’m also attaching as a reply, my scratch sheet showing some calculations I made. I decided to leave the integral I marked L as an interesting exercise for the spunky reader. 🤓 Note that it goes from 0 to ♾️.
30 Days of Feynman—Day 20
This is a continuation of Day 19. All 3 integral formulas presented below may be derived using only the results from Day 18, plus 2 trig reduction formulas.
For days 18-20, assume a>0.
Contour integration may also be used if you have extra paper.
30 Days of Feynman—Day 16
Solutions involving differentiating under the integral sign.
The solution process includes a separable first order differential equation. It also requires knowledge of the Gaussian integral.
A new take on the “Gifted” integral. Seven year old prodigy, Mary, noted that the integral, as originally written (with a positive exponent), would not converge. After correcting the integral, she solved it using a double integral in polar coordinates. I use the gamma function.
30 Days of Feynman—Day 17
Solutions involving differentiating under the integral sign.
Here’s two integrals for the price of one. The solution process includes a separable first order differential equation, and requires knowledge of Euler’s formula and the Gaussian integral.
Here’s yet another interesting (though obviously contrived 🧐) integral involving the golden ratio. This integral shows up repeatedly on X. Because it was so quick to evaluate, I’m also including the calculation of arcsinh(1/2) from scratch.
30 Days of Feynman—Day 15
Solutions involving differentiating under the integral sign.
This solution requires knowledge of Euler’s formula and the gamma function. It also requires some working knowledge of the relationship between trig functions and inverse trig functions.
30 Days of Feynman—Day 23
Like Days 21 and 22, here’s an example of integration under the integral sign.
For further information, see the linked article, which references a 100 year old advanced calculus book:
Lunchtime (in Texas) integral. Note that the integral features the sum of cubes in the denominator. I employed the same solution method as I have been using the past few days, but first try it yourself.
40 Days of Feynman—Day 24
I’ve renamed the series. Here’s yesterday’s integral. I made a couple of corrections. I will be using the results in Day 25.
In the solution, I used the same concept as in Days 21-23, but just presented it differently.
30 Days of Feynman—Day 11
For 30 days, I’m publishing solutions involving differentiating under the integral sign.
In a reply I attached to this post, I evaluated the intermediate integral using the tangent half-angle substitution, also known as the Weierstrass substitution.
I slightly modified an integral recently posted here by
@Twins65489923
. As replies to this integral I will explain some of the steps. The solution requires passing knowledge of the binomial series, the Dirichlet beta function, and the definition of Catalan’s constant (G).
40 Days of Feynman—Day 26
This integral formula also uses the results from Day 24. You may, of course, generalize even further, according to the results from Day 25, if you have the patience to do a nasty alphabet soup derivative.
I threw in 2 examples using the formula.
30 Days of Feynman—Day 14
Solutions involving differentiating under the integral sign.
This solution is longer only because I included many more details. I am attaching my notes as replies 1 and 2, as referenced in the solution. You may find them interesting.
Both integrals are pretty easy, especially if you’ve been following the integrals I’ve been posting here for the past week. I(a) is from Spivak’s book; I(b) is related.
This is an integral
@drpkmath
originally posted here, and on his YouTube channel, with a different solution method.
Using the Feynman trick, I introduce a parameter on the first line of the solution, and then differentiate under the integral sign with respect to that parameter.
This problem was inspired by a problem posted by
@MYosefnia
earlier today. I modified it somewhat to suit my tradition of posting gamma function-related problems over the weekends. It also fits in nicely with my recent Euler’s formula theme.
In keeping with the theme of the past few days, here’s another integral solution employing the Feynman trick, the geometric series, and the Dirichlet eta function. For further info, see the solutions I’ve posted the past few days.
I evaluated this interesting integral using the gamma function and Dirichlet eta function. In keeping with the Greek theme, the eta function is the alternating version of the Riemann zeta function. Typo corrected.
This is a variation of an integral presented by
@Twins65489923
a couple of weeks ago. My solution features Laplace transforms (the cursive L), the beta/gamma function, and an interesting manipulation of Euler’s reflection formula. I hope everyone’s having a great day. 🌞
Here’s another integral from Spivak’s Calculus (a book that’s freely available online).
@Diarytells
presented a form of this integral yesterday. Spivak’s integral is easy in comparison.
Today’s integral is from the excellent account of
@MYosefnia
. It requires knowledge of the beta function and Euler’s reflection formula. All required info can be found in Wikipedia.
A foundational integral, solved using formulas previously derived here. There are, of course, other ways to express the anti-derivative, depending on how you solve the problem.
Here’s the same integral I evaluated yesterday via contour integration. This time, I use the tangent half angle (Weierstrass) substitution. Because of the singularity, it’s not a straightforward process. |a|>1
30 Days of Feynman—Day 18
Solutions involving differentiating under the integral sign.
An integral formula 2-fer. The solution process includes a simple second order differential equation, and requires knowledge of the Dirichlet integral.
Often done via contour integration.
I evaluated the given integral using the Feynman trick. Somewhere in my archives is my evaluation of this integral using complex analysis. If I find it, I will post it. Perhaps
@unsolved_equ
will show us how to evaluate this using Laplace/inverse Laplace transforms.
I’m presenting this solution as a response to an integral solution I saw on YouTube earlier today. It may be helpful to folks to see two different ways to evaluate the same integral.
I couldn’t let the weekend go by without evaluating an integral. The imaginary number i appears a couple of times in the integrand, but it should be no cause for concern. I assume some knowledge of the Gaussian integral.
Saturday morning beta/gamma function. Integrals I and J are integrals that Cipher evaluated using other techniques. See Cipher’s solutions in the replies.
30 Days of Feynman—Day 22
Like Day 21, here’s an example of integration under the integral sign (not differentiation). This is the same integral from Day 4 (using pi and e instead of a and b). I’m showing the Day 4 solution in the replies, so that you may compare the methods.
Two more integral formulas derived by using Euler’s formula. Of course, the alternative to this approach is integration by parts, or using trig identities derived last week.
Posted by my friend,
@muntassirs
, in response to a private query. I encourage all X mathematicians to post their full solutions to interesting integrals (and other math problems), so that we can all learn from each other. We’re all at different points in our math(s) journey.
30 Days of Feynman—Day 13
Solutions involving differentiating under the integral sign.
Even though the imaginary number i appears in the integrand, the integral evaluates to a real number. Note that the parameter is introduced in two places. i*arctanh(x)=arctan(ix).
Corrected
Same themes as explored in solutions I’ve posted the past few days. This post assumes some knowledge of the Dirichlet beta function. But everything you need to know about that function to evaluate this integral is contained in the Wikipedia article that I’ve attached as a reply.
Here’s how to do step 7 for the integral solution I posted this past Friday. The solution to this sub-integral requires a nodding acquaintance with the beta function, properties of the gamma function (including its relationship to the beta function), and combinatorics.
30 Days of Feynman—Day 21
As with Days 18 and 19, the formula for the integral below may be derived using only the results from Day 18.
The new wrinkle here is the integration under the integral sign, rather than differentiation. Everything else is the same.
A weird limit problem that
@CPierre67
and I discussed over lunch today. This is a postscript to the infinitely-nested radical that we integrated last week. Your thoughts 💭
Cipher just posted his solution to the following integral on his terrific YouTube channel. He evaluated the integral using Laplace transforms. I found it much easier to use a property of the cosine integral function. If you find another method, please post it below.
This is a solution to a problem recently posted by
@robolsmath
, who is worthy of a follow for math folks. The solution involves a trick of sorts: convert a difficult integral into a much easier integral plus an integral that always evaluates to 0.
More weekend fun with the gamma function.
Using the formula I derived (see reply to this post), and only properties of the gamma function, I evaluated two integrals (
#1
should look familiar).
gamma(x)gamma(1-x)=pi/(sin(x.pi)).
Thus, gamma(1/2)=root(pi).
gamma(x+1)=xgamma(x).
Here’s the derivation for items number 7 and 8 from the list of Laplace transforms that I attach as a reply. An interesting exercise for the reader will be to derive all of the other ones, or at least one of the other ones. 🤓
The same themes that I’ve covered the past few days run through this integral. Here, I provide somewhat fewer details, but I guess you get the general idea. Have a nice week.
Another irenic integral from the account of
@MYosefnia
, modified slightly to discourage use of the Barnes G-function. My solution features the Feynman trick, well-known properties of the gamma function, and a previous result, which I attach as a reply.
A quick one, if you know the trick. The trick is explained in one of the replies. The other reply contains the video solution (with a slightly different method) by Cipher.
Is there a “best” way to write an inverse trig function, like tan? If not, what is your preference a, b, c, d, e, and/or f from the list below? Which one do you like least? State your reasoning. If there’s one that’s not on my list that you either like or dislike, please add it.
My solutions to two integrals related by Euler’s formula. Prof PK (
@drpkmath
) presented a full solution to one of these on his YouTube channel earlier today.
My solution to a
@Diarytells
integral from yesterday. I use two common tricks: multiplying by a form of 1, and adding a form of 0 to the numerator to get everything to work out. Finally, I use a (somewhat) well-known integral formula at the very end.
Keeping in practice with the tangent half-angle substitution, sometimes erroneously referred to as the Weierstrass substitution. Integral under the account of
@sonukg4india
.
Another integral solution using a geometric series, the gamma function, and the Riemann zeta function. From
@MYosefnia
.
Note: gamma(8)=7!=5040,
Zeta(8)=pi^8/9450
Corrected. Left out 2 details.
Here’s an integral that you can solve in your head (no scratch paper) in less than 10 seconds, if you’ve been following my account recently. If you have another solution method, please share it. BTW, I use “G” for Catalan’s constant, rather than C.
Here’s the first two integrals from the list posted a couple of days ago. I’m assuming that the first integral was intended to be an improper integral from zero to infinity.
If you liked the integral,
@Diarytells
posted yesterday, you may also like this one. Note that we’re mapping to the unit circle in the complex plane in the negative direction. Otherwise, your sign will be wrong.