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I was trying to evaluate $ {\int_0^{\infty}\frac{\cos (x)}{1+x^2}}\text{d}x$ by Feynman's trick

It has been solved here, but I have a specific question at one of the steps while doing it. (Not a duplicate) (My question is about why I get 2 different results when solving it in 2 different methods in a particular step.)

So I defined $I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$ and arrived at $I(t)= I''(t)$

so this means

$I(t) = \alpha e^t + \beta e^{-t}$

Method 1:

$I(0) = \alpha + \beta $

and from the integral $I(0) = \int_0^{\infty}\frac{1}{1+x^2}\text{d}x$ = $\frac{\pi}{2}$

so $\alpha + \beta = \pi/2 \tag1 $ and $I'(t) = \alpha e^t -\beta e^{-t}$, so $I'(0) = \alpha -\beta $

from the integral $I'(t)= \int_0^{\infty}\frac{-x \sin (xt)}{1+x^2}\text{d}x$

so $I'(0)= 0 $ $\alpha - \beta = 0\tag2$

By (1) and (2) we get $$\boxed{I(t) = \frac{\pi}{4}e^t + \frac{\pi}{4} e^{-t}}$$

Method 2:

$I(0) = \alpha + \beta $

and from the integral $I(0) = \int_0^{\infty}\frac{1}{1+x^2}\text{d}x$ = $\frac{\pi}{2}$

so $\alpha + \beta = \pi/2 \tag3 $

$I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$

using integration by parts,
$$u=\frac{1}{{{x}^{2}}+1}\quad,\quad dv=\cos tx$$ we have $$I(t)=\frac{\sin (t x)}{t({{x}^{2}}+1)}\left| \begin{matrix} \infty \\ 0 \\ \end{matrix} \right.+\frac{2}{t}\int_{0 }^{+\infty }{\frac{x\sin (t x)}{{{({{x}^{2}}+1)}^{2}}}}\,dx $$ as a result $$I(t )=\int_{0}^{\infty }{\frac{2}{t} \frac{x\sin tx}{{{({{x}^{2}}+1)}^{2}}}\,}dx \,$$

As t tends to infinity we can see I(t) will tend to 0, which should mean $\alpha$ is 0, or else it will exponentially increase.

So now we get $$\boxed{I(t) = \frac{\pi}{2e^t}}$$

What's going wrong? Both seem valid and correct. This seeming loophole has been annoying me for a long time

Rick
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    How did you get $I=I’’$? – Paul Dec 30 '17 at 15:58
  • how do you justity that $I''$ even makes sense? the integral in question seems to diverge – tired Dec 30 '17 at 16:09
  • I was looking at this solution https://math.stackexchange.com/a/1841104/ – Rick Dec 30 '17 at 16:11
  • go for the answer of @Aryabhata it is much simpler to justifiy rigorously – tired Dec 30 '17 at 16:13
  • @GuyFsone, no it isn't. I have a specific question in a step while calculating it, not the whole calculation – Rick Dec 30 '17 at 16:17
  • @Rick Both trial you did are given in the answers of the given link – Guy Fsone Dec 30 '17 at 16:19
  • @tired: The integral $I(t)$ does not diverge for $t>0$, because it is the limit of an alternating series with terms going to zero. – user21820 Dec 31 '17 at 04:05
  • Can the people who voted to close my question as a duplicate please tell how the linked one answers it? I looked though all 9 answers there and none of them discusses this particular "loophole". – Rick Dec 31 '17 at 04:28
  • I think Paul's comment is right; you did not prove that $I'' = I$. – user21820 Dec 31 '17 at 04:41
  • @user21820 It's proved in that linked question, but I did it like $I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$, so $I'(t)= \int_0^{\infty}\frac{-x \sin (xt)}{1+x^2}\text{d}x$ and $I''(t)= \int_0^{\infty}\frac{-x^2 \cos (xt)}{1+x^2}\text{d}x$. From this $I(t) - I''(t) = \int_0^{\infty}\cos (xt)\text{d}x$. This answer said the integral from $-\infty$ to $\infty$ is 0, since it's an even function 0 to $\infty$ should be 0 too... – Rick Dec 31 '17 at 04:53
  • That is nonsense. That last integral does not exist. Don't anyhow believe what this or that answer says. – user21820 Dec 31 '17 at 05:17
  • @user21820 Ohhhh..basically this whole "paradox" arises because of that nonsensical step over there...I was getting confused for nothing. Thanks a lot and happy new year :) – Rick Dec 31 '17 at 14:27
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    You're welcome and same to you! =) – user21820 Dec 31 '17 at 14:37

1 Answers1

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You are using integration by parts to get $$I(t)=\frac{\sin (t x)}{t({{x}^{2}}+1)}\left| \begin{matrix} \infty \\ 0 \\ \end{matrix} \right.+\frac{2}{t}\int_{0 }^{+\infty }{\frac{\sin (t x)}{{{({{x}^{2}}+1)}^{2}}}}\,dx$$ with $$u=\frac{1}{{{x}^{2}}+1}\quad,\quad dv=\cos (tx)$$ When you differentiate the expression for u , you get $$du=\frac{-2x}{{({x}^{2}}+1)^2}\quad\quad$$ Apparently you missed the x in the numerator of the fraction $$du=\frac{-2x}{{({x}^{2}}+1)^2}\quad\quad.$$

Mohammad Riazi-Kermani
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