This Putnam Integral Has a Ridiculously Simple Answer

Added: 2026-05-11

[00:00:00]Okay, here we have another pundam integral. So we have integral from 0 to infinity t ^ of1 /2 * e ^ of -1,985 * t + 1 / t dt. At the first glance, this integral looks strange. We have t + 1 / t and then we have the strange factor of t to the power of 1 /2.

[00:00:23]But when we evaluate this, we're not going to evaluate this directly, but we'll make a copy of it. add a two together then suddenly that will become Gaussian integral. So notice how this t + 1 / t okay this is symmetric under the transformation of t as 1 / t. So that's why we can make a substitution t as say 1 / u right.

[00:01:01]So let's make a substitution t s as 1 / u.

[00:01:06]Okay. Then your dt is the same as -1 / u^2 du.

[00:01:14]Then at the same time we have that t to the power of 1 /2. So using this t the power of -1 / 2. Okay. This is then going to be 1 / u 2 ^ of - 1 /2. So that this is just the same as u ^ of 1 / 2.

[00:01:33]Okay. Now let's just talk about t + 1 / t.

[00:01:39]So t + 1 / t is then going to be then the same as 1 / u plus u.

[00:01:48]And let's just talk about lower bound and the upper bound. Right? So the lower bound of the t was zero. Upper bound of the t is infinity.

[00:01:56]So when t is going to zero then your u is going to infinity. Okay. Then when your t is going to infinity then your u should be going to now zero.

[00:02:12]Okay. So that's why let me just call this integral as the i first.

[00:02:17]and your I is the same as integral from infinity to zero using the substitution. Then we have u to the power of 1 / 2 that times e ^ of -1,985.

[00:02:36]Okay. Then that times u + 1 / u.

[00:02:41]Okay. And your dt was 1 / u ^2 du. So times -1 / u ^2 and we have du.

[00:02:51]But then again if you pull this negative sign outside of this integral then we can switch this lower bound to the upper bound by pulling another negative sign.

[00:02:59]The negative negative become positive.

[00:03:00]So we can just rewrite this integral I as integral from 0 to infinity of u ^ of 1 /2 times let me just put 1 / u ^2 first and then that times ^ of -1985 * u + 1 / u and then we have du.

[00:03:26]So u ^ of 1 /2 * 1 / u ^2 that is the same as u ^ of -3 /2 right. So using that this is the same as integral from 0 to infinity that of u ^ of -3 / 2 and everything else has to be just the same time e ^ of -1,985 * u + 1 / u. And then we have du.

[00:04:01]And if you turn this u to the t, then your integral i is also the same as integral from 0 to infinity of turn this u to t. t ^ of -3 /2* e ^ of -1,985 time + 1 / t. Then we have dt.

[00:04:28]This has to be the mirror version of your integral I. So so far you have integral let me just write this down again. I was integral from 0 to infinity that of t ^ of -1 / 2 * e ^ of -1,985 * t + 1 / t dt.

[00:04:49]Then there's mirror version of this integral I. I is also the same as integral from 0 to infinity of e ^ of 3 /2 * e ^ of -1,985 * the same t + 1 / t and then we have dt.

[00:05:10]Like I said, I'll be adding these two up, right? That is it has to be the same as 2 * i. It is integral from same 0 to infinity and then since we have the same term about the e right so we can just add those two up parenthesis t to the power of -1 / 2 plus t ^ of -3 / 2 close your parenthesis that times e ^ of -1,985 that times t + 1 / t and then we have dt and it looks like we can work on this parenthesis t to the^ of 1 /2 + t ^ of -3 /2. So let's just work on this parenthesis part.

[00:05:59]So t ^ of1 /2 + t ^ of -3 /2. This is going to be then the same as first t the^ of 1 /2. We can rewrite this as t over t ^ of 3 /2. Second term it is the same as just a 1 / t ^ of3 over 2.

[00:06:20]So we can just combine this as t + 1 over t ^ r3 / 2.

[00:06:30]Okay. So using this we can just rewrite this 2 * i.

[00:06:37]2 I has written the same as integral from 0 to infinity of t + 1 over t ^ of 3 / 2 times the same e ^ of -1,985 * t + 1 / t. Then we have dt.

[00:06:57]Still the integral is looking ugly. So time to use another substitution. That substitution is x is square root of the t minus 1 / roo<unk> of the t. And there's a reason why if you square both side then something amazing would happen. So if you go ahead and do this then x² is going to be then the same as t then minus q + 1 / t. So that's why t + 1 / t is the same as x^2 + 2.

[00:07:30]So we'll be using this and at the same time this e ^ of -1,985* t + 1 / t that's going to be just the same as e ^ of1,985 times then x² + 2.

[00:07:53]Okay, then let's just talk about dx, right?

[00:07:56]So your x was now this x was roo<unk> of the t minus 1 / roo<unk> of the t. So we can just rewrite this as x is the same as t ^ of 1 /2 and then minus t ^ of -1 /2. And let's get derivative on the left and right hand side. So if you go ahead and do this then your dx is the same as let me just pull this one over two out right. So 1 / 2 parentheses t to the power of -1 / 2 and then minus - plus and we already pulled this one over two out. So we should have t ^ of -3 / 2 then we have dt.

[00:08:43]Okay. So using this but then again t to the power of 1 /2 plus t ^ of -3 /2 we already calculated this right? it was t + 1 / t ^ of 3 over 2. So we already have an expression for this part that was t + 1 over t ^ of 3 /2.

[00:09:04]And at the same time then d x over dt. dx over dt is then going to be the same as 1 / 2 * t + 1 over t ^ of 3 /2.

[00:09:22]Okay. So that is why now we can say t + 1 over t ^ of 3 / 2 dt is just equal to 2dx.

[00:09:38]So now we have everything. So let's just talk about lower bound and the upper bound right. So the lower bound is when t is going to zero your x is going to negative infinity and when your t is going to infinity of course x is going to positive infinity. So that's why your 2 * i. So using this 2 * I would be rewritten as integral from negative infinity to positive infinity. Okay. Then that of 2 * e ^ of -1 985. Okay. Then that times x² + 2.

[00:10:29]Okay. Okay, then we have just a dx.

[00:10:36]And if you're pulling constant part from the exponent, this is the same as 2 * e ^ of double this 1,985.

[00:10:45]It is going to be -3,970 the times integral from negative infinity to infinity of the same e to the power of -1,985 the times just the x² then we have dx now let's take a look at this integral right so if you're looking at this integral integral from negative infinity to infinity e ^ of1985 x² dx X looking exactly like this Gaussian integral. Right? So the Gaussian integral in general if you have integral from negative infinity to infinity of e ^ of a x² dx the result of it is just going to be the same as square root of pi / a.

[00:11:35]If your a is positive I already evaluate this gaussian integral from one of my previous videos.

[00:11:42]So if you haven't watched it make sure to watch it. So in our case your a has been the same as okay this 1,985 right so in this case a has to be 1,985 so that is why the result of this part is going to be the same as square root of pi over 1,985 okay so that is why that's two times integral I is the same as 2 * e ^ of -3,970.

[00:12:20]That times square<unk> of pi over 1,985.

[00:12:26]We're looking for the value of the integral i. Right? So that's why we can just divide both side by two. Then your integral i is the same as e ^ of -3,970.

[00:12:38]the times square<unk> of pi over 1,985.

[00:12:47]So this is the answer for the question.

[00:12:52]Okay, it's a pretty interesting pond integral. How amazing.