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The following integral i was trying to evaluate: $$\int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx.$$ ,what i did was substituting pi/x = 1/t in the second integral , which converts to first integral which tells integral is zero ,but answer is ln(pi) , whats the fault that i wanted to know .

WizardMath
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    "which converts to first integral which tells integral is zero". You can't do that. You need to perform the u-sub on BOTH integrals and not just one and then "combine" the results.Check DESMOS to plot both functions – imranfat Mar 15 '21 at 00:16

2 Answers2

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It sounds like what you did was to write $$ \int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx = \int_0^\infty \sin(1/x)\, dx - \int_0^\infty \frac{\sin(\pi/x)}{\pi} \,dx $$ You then made a $u$-substitution on the second term $u = x/\pi$, in which case the integrals are the same and "should" cancel.

The problem is that both of these improper integrals are divergent. Technically what we have here is $$ \lim_{R \to \infty} \left[\int_0^R \sin(1/x)\, dx - \int_0^R \frac{\sin(\pi/x)}{\pi} \,dx \right] $$ but since both integrals approach $\infty$, this is a limit of the form $\infty - \infty$ and is not necessarily zero.

If we do the $u$-substitution for the proper integrals, it becomes instead $$ \lim_{R \to \infty} \left[ \int_0^R \sin(1/x)\, dx - \int_0^{R/\pi} \sin(1/u) \,du \right] = \lim_{R \to \infty} \int_{R/\pi}^R \sin(1/x) \, dx $$ which is more amenable to limit-taking. (In particular, you might try substituting $w = 1/x$ at this point.)

Michael Seifert
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  • I got the pt of infinity - infinity form Sir , but dont we need to consider the R as R tends to 0 , bc thats we dont know in both integrals bc both tends to infinity- infinity ,if R tends to infinity we know its exactly 0-0 isnt ? [ The value at those R , i am referring ] – WizardMath Mar 15 '21 at 00:02
  • I meant to say lower one should be R , not the above one as valye at R tends to infinitt we know both values goes to zero – WizardMath Mar 15 '21 at 00:07
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    @WizardMath: I'm not sure I entirely understand what you're asking. But my point in the above is that the proper integrals in Eq. (1) above are not the same, and that both of them tend to infinity individually as $R \to \infty$. – Michael Seifert Mar 15 '21 at 00:16
  • $$ \lim_{R \to \0} \left[\int_R^\infty \sin(1/x), dx - \int_R^\infty\frac{\sin(\pi/x)}{\pi} ,dx \right] $$ Sir this i meant , bc dont we always for solving improper integrals take the limit where the function is not defined , like here R = 0 one bc sin(infinity) , am i right ? – WizardMath Mar 15 '21 at 00:40
  • And Sir i got your pt of how u proved that its not zero , i just meant the above thing which we usually do when solving improper integrals , and also Sir $$ \lim_{R \to \infty} \int_{R/\pi}^R \sin(1/x) , dx$$ after doing w=1/x substiutions what we need to do after this ? as we get integral from pi/R to R -(sinw/w^2) dw, i dont get idea after this – WizardMath Mar 15 '21 at 00:55
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    Concerning your question about the lower bounds: I suppose you're right that technically we need to be taking a limit as both the upper bound goes to infinity and the lower bound goes to zero. I'm actually not sure how to argue that the limit of the lower bounds goes to zero; I'll have to think about that further. Concerning how to do the integral once you've substituted with $w$: I would expand the integrand in a Taylor series and integrate it, but I'm a physicist and that technique might not be rigorous enough for a "real" math class. – Michael Seifert Mar 15 '21 at 02:45
  • Yeah alright Sir tell after you thought about it and Sir that series expansion method looks nice , if u can post it as as a second answer i can upvote that too. I like to know that approach :) . I also wanna be a physicist so i have np in that :) – WizardMath Mar 15 '21 at 03:45
  • @WizardMath: Lower limit does not create a problem as you can define the function in any manner at $0$. The final answer is easily obtained by adding and subtracting $1/x$ from integrand. – Paramanand Singh Mar 16 '21 at 04:55
  • I meant consider the situation of like for this integral : integral from 0 to 1 of( 1/x )dx , wont we take limit at lower value so that first we get the integrated function and then put the lower value limit to it ? And yeah Sir can u share your method too if u dont have problem ? – WizardMath Mar 16 '21 at 05:00
  • +1 for the simple approach. Also you have given a very nice explanation about why it is wrong to split the integral as difference of two integrals. – Paramanand Singh Mar 16 '21 at 05:05
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    @WizardMath: $\sin(1/x)$ is different from $1/x$ in a fundamental way. It is bounded near $0$. The improper integrals are used when integrand is unbounded or interval of integration is unbounded. – Paramanand Singh Mar 16 '21 at 05:06
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    @WizardMath: I don't want to write a separate answer which is based on this. Consider the last integral at end of this answer and write it as $\int_{r/\pi} ^{r} (\sin(1/x)-(1/x)),dx+\int_{r/\pi}^r(1/x),dx$. The second integral is $\log\pi$ and first one tends to $0 $. Why? Because using mean value theorem you can write it as $(r-(r/\pi))(\sin(1/c)-(1/c))$ where $r/\pi<c<r$. When $r\to\infty $ this tends to $0$. – Paramanand Singh Mar 16 '21 at 05:12
  • Wow , really neat Sir , thx a lot for sharing it :) – WizardMath Mar 16 '21 at 05:16
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Integrate by parts to get $-x\dfrac{\sin\left(\frac{{\pi}}{x}\right)}{{\pi}}+x\sin\left(\dfrac{1}{x}\right)+\operatorname{Ci}\left(\dfrac{{\pi}}{x}\right)-\operatorname{Ci}\left(\dfrac{1}{x}\right)$

$(1).$ As $x\to \infty,$ the first two terms combine to give $-1+1=0.$ As $x\to 0,$ all four terms tend to $0$. Now,

${\displaystyle \operatorname {Ci} (x)=-\int _x^\infty \frac {\cos t}{t}}dt$ so ${\displaystyle \operatorname {Ci} (\pi/x)=-\int _{\pi/x}^\infty \frac {\cos t}{t}}dt$ and $\displaystyle\operatorname {Ci} (1/x)=-\int _{1/x}^\infty \frac{ \cos t}{t}dt$

and therefore

$(2).\ \operatorname{Ci}\left(\dfrac{{\pi}}{x}\right)-\operatorname{Ci}\left(\dfrac{1}{x}\right) = \displaystyle \int _{1/x}^{\pi/x} \frac {\cos t}{t}dt=\int _{1/x}^{\pi/x}\left(\frac{1}{t}-\frac{t}{2}+O(t³)\right)dt=$

$\ln(\pi/x)-\ln(1/x)+O(1/x^2)=\ln \pi+O(1/x^2)$

To finish, let $x\to \infty$ in $(2)$ and combine the result with $(1).$

imranfat
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Matematleta
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  • Wow this is really nice Sir , one thing i wanna ask how about doing it with feynman techinique , is it possible if we take that pi in the sin term in second integral as a variable a ? @Matematleta – WizardMath Mar 15 '21 at 03:47
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    The Feynman teaching method? I am not sure what you mean. If you let a vary, then you get a function $a\to \ln a$ – Matematleta Mar 15 '21 at 18:21
  • I meant feynman techinque of differentiation under integral sign ,then integrate it back . Yeah can u share how u got answer as lna @Matematleta Sir – WizardMath Mar 16 '21 at 00:08
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    Switching the operations is valid because the McLaurin series for $\cos t/t$ is unifomly convergent on $[1/x, \pi/x]$ – Matematleta Mar 16 '21 at 00:59
  • Wym by "switching the operations " Sir , arent we differentiating it ? – WizardMath Mar 16 '21 at 04:25
  • No, we expand the integrand in a series and integrate term by term. – Matematleta Mar 16 '21 at 15:27
  • I see but i was talking of newtons lebinitz differentuation under integral sign Sir – WizardMath Mar 16 '21 at 15:29
  • Why would you want to do that? L'Hospital's Rule on the two indeterminate forms in the itegrand? I do not think that will work You will get a circular argument. – Matematleta Mar 16 '21 at 17:07