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for this integral, I plan to solve it by using residues, but it seems when dealing with the integral over the upper semi-circle, it doesn't converge. Could you give some help, thank you very much! $$\int_{-\infty}^{\infty}\frac{\sin(x-\frac{a}{x})}{(x+\frac{1}{x})}dx$$

Asaf Karagila
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MathFail
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  • Can you show your execution of your "plan"? The work behind your conclusion? – amWhy Oct 11 '21 at 15:17
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    I have a solution, but without using residues. I can answer, if you are interested. – Laxmi Narayan Bhandari Oct 11 '21 at 15:27
  • Yes, could you please give some hint without using residues? Thank you :) – MathFail Oct 11 '21 at 16:16
  • Ah, I think I know how to solve it now, but it still needs to calculate the residues. So could you share your method please? @LaxmiNarayanBhandari – MathFail Oct 11 '21 at 16:31
  • I don't know if it will help, but at first glance I thought about this: https://mathworld.wolfram.com/GlassersMasterTheorem.html – Teruo Oct 11 '21 at 22:37
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    thank you, and I have solved it by using residues :) @P.TeruoNagasava – MathFail Oct 12 '21 at 00:49
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    Please, add in the post the solution with residues. – Claude Leibovici Oct 12 '21 at 05:24
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    You can argue that the value of the integral is $\Im \left(2 \pi i \operatorname{Res} \left[\frac{z e^{iz}e^{-ia/z}}{1+z^{2}}, i \right] \right) = \frac{\pi}{e^{a+1}}$ by first showing that $\left|e^{-ia/z}\right| \le 1$ in the upper half plane if $a \ge 0$. Then use Jordan's lemma. – Random Variable Oct 12 '21 at 08:28
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    It is interesting to mention that the approach proposed by @Random Variable allows to evaluate a more general integral ($a,b>0; c\in R$): $$I(a,b,c)=\int_{-\infty}^\infty\frac{e^{ibx-\frac{ia}{x-c}}}{x+\frac{1}{x}}dx=\pi i,e^{-b-\frac{a}{1+c^2}}e^{-i\frac{ac}{1+c^2}}$$ $$I_1=\int_{-\infty}^\infty\frac{\sin,(bx-\frac{a}{x-c})}{x+\frac{1}{x}}dx=\pi,e^{-b-\frac{a}{1+c^2}}\cos\Big(\frac{ac}{1+c^2}\Big)$$ – Svyatoslav Oct 12 '21 at 09:59
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    @Svyatoslav The last integral isn't possible to evaluate with my approach, though. Complex analysis is much more powerful for such integrals. – Laxmi Narayan Bhandari Oct 13 '21 at 02:25
  • @MathFail, it's quite not nice to remove the question by vandalising it. Some have put effort (and quite a lot!) to answer your question and which most certainly solved the problem and helped you. If so accept it and put back the question as it was. You see, this site not just for individuals, it is also for a Community. So this could help someone else too – Vega Nov 22 '21 at 15:58
  • @Vega No, it is not removed, it is closed by system. My post was to ask if it can be solved by contour integral, nothing else... I see, I will accept Laxmi's solution then :) – MathFail Nov 23 '21 at 12:10
  • You removed the question text by yourself, that's what I was saying. You can still improve your question and vote to reopen – Vega Nov 23 '21 at 12:25

1 Answers1

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I am not going to use residues for this. This answer is going to abuse differentiation under the integral sign.

We start by noting that the integrand is even and thus,

$$I (a)=2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}{x+\frac1x}\,\mathrm dx$$

Throughout the solution, I will be assuming that $a \geq 0$.

Differentiating w.r.t. $a$, $$I'(a) = -2 \int\limits_0^\infty \frac{\cos\big(x-\frac ax\big)}{x^2+1}\,\mathrm dx $$ Again differentiating, $$\begin{align}I' '(a) &= -2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}{x(x^2+1)} \,\mathrm dx \\ &= -2 \int\limits_0^\infty\frac{\sin\big(x-\frac ax\big)}x-\frac{x\sin \big(x-\frac ax\big)}{x^2+1}\,\mathrm dx\end{align}$$

Using the substitution $x\mapsto \frac ax$, we can easily prove that $ \int\limits_0^\infty\frac{\sin(x-\frac ax)}x=0$. Thus,

$$I' '(a) = 2 \int\limits_0^\infty \frac{x\sin\big(x-\frac ax\big)}{x^2+1}\,\mathrm dx= I(a) $$

Now, we have a differential equation, cool!

$$\begin{align}I(a) &= I' '(a) \\ \implies I(a) &= k_1 e^a+k_2 e^{-a} \end{align}$$ For evaluating the constants, we have the initial values $(I(0), I'(0))=\big(\frac\pi e, -\frac\pi e \big)$. Using this, we get $(k_1,k_2)=\big(0,\frac\pi e\big)$.

Thus, we conclude that $$I(a) = \int\limits_{-\infty}^\infty\frac{\sin\big(x-\frac ax\big)}{x+\frac1x}\,\mathrm dx = \frac\pi {e^{a+1}} $$

Laxmi Narayan Bhandari
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    This is more than impressive ! – Claude Leibovici Oct 12 '21 at 05:17
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    A nice approach (+1). Is it possible to evaluate $\int\limits_{-\infty}^\infty\frac{\sin\big(bx-\frac ax\big)}{x+\frac1x},\mathrm dx (a,b>0) $ in this way? – Svyatoslav Oct 12 '21 at 06:13
  • @Claude Thanks a lot! – Laxmi Narayan Bhandari Oct 12 '21 at 07:40
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    @Svyatoslav The procedure will stay the same. Only the zeroing substitution will change to $x\mapsto \frac a{bx}$. – Laxmi Narayan Bhandari Oct 12 '21 at 07:56
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    $$I(a,b)= \int\limits_{-\infty}^\infty \frac{x\sin\big(bx-\frac ax\big)}{x^2+1},\mathrm dx$$ In the same way as you did (and using your substitution $x\mapsto \frac a{bx}$) $$I(a,b)=k_1e^a+k_2e^{-a}; ,I(a=0,b)=\pi e^{-b}; ,I'_a(a=0,b)=-\pi e^{-b}$$ $$\Rightarrow I(a,b)=\pi e^{-(a+b)}$$ – Svyatoslav Oct 12 '21 at 08:29
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    @Svyatoslav Noice. – Laxmi Narayan Bhandari Oct 12 '21 at 08:38
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    Wow, this method is really fantastic!! Thank you so much! @LaxmiNarayanBhandari – MathFail Oct 12 '21 at 13:02
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    This trick $x->a/x$ let the integral $\int_0^\infty \frac{\sin(x-\frac{a}{x})}{x}dx$ vanish, because inside the sine function it is subtraction, right? It seems if we change the numerator to $\sin(x+\frac{a}{x})$, then this trick seems not work? In this case, must we compute the residues the only way? @LaxmiNarayanBhandari – MathFail Oct 12 '21 at 13:27
  • You guessed it right. This trick is only useful in this particular case with sine and subtraction between the two fractions. Otherwise, the only way would be to actually compute the integral. The fun fact is that if you evaluate the original, but with cosine instead of sine, you will get exactly the same result. Must try it! @MathFail – Laxmi Narayan Bhandari Oct 13 '21 at 02:01
  • @Math for the arrow sign I have used, use \mapsto . This will give $\mapsto $. – Laxmi Narayan Bhandari Oct 13 '21 at 02:02
  • @Svyatoslav Do the same integral, but with cosine instead of sine. – Laxmi Narayan Bhandari Oct 13 '21 at 02:03
  • $$\int_{-\infty}^{\infty}\frac{\cos(x-\frac{a}{x})}{(x+\frac{1}{x})}dx =0$$ - if we integrate $(-\infty; \infty)$ (the integrand is odd). This is also visible from the general case: $$I_2=\int_{-\infty}^\infty\frac{\cos,(bx-\frac{a}{x-c})}{x+\frac{1}{x}}dx =\Re ,\pi i,e^{-b-\frac{a}{1+c^2}}e^{-i\frac{ac}{1+c^2}}=\pi,e^{-b-\frac{a}{1+c^2}}\sin\Big(\frac{ac}{1+c^2}\Big)$$ $$I_2=0 ,(\text{at} ,c=0)$$ For the integral $(0; \infty)$ it is more complicated; for example, $\int_{0}^\infty\frac{\cos,(x-\frac{a}{x})}{x}dx\neq0$ – Svyatoslav Oct 13 '21 at 02:54
  • @Svyatoslav I used my method, but on the limits $(0,\infty)$, and I got the same result as the integral with sign. Try my method. – Laxmi Narayan Bhandari Oct 13 '21 at 03:56
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    $$I(a)=\int_{-\infty}^{\infty}\frac{\cos(x-\frac{a}{x})}{(x+\frac{1}{x})}dx$$ $$I''{aa}=I(a)-\int{0}^\infty\frac{\cos,(x-\frac{a}{x})}{x}dx$$ But the second term is not zero: after the turn in the complex plane ($x=it$) $$\int_{0}^\infty\frac{\cos,(x-\frac{a}{x})}{x}dx=\int_0^\infty\frac{e^{-t-\frac{a}{t}}}{t}dt=2K_0(2\sqrt a)$$ - if I made all correctly (https://math.stackexchange.com/questions/403576/evaluating-int-0-infty-frace-t-fracxtt-dt?noredirect=1) – Svyatoslav Oct 13 '21 at 14:44
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    @Svyatoslav Probably you are correct. I missed that my method would work on $\int_0^\infty \frac{\cos(x-\frac ax)}{x^2+1}, dx $. Nice work. – Laxmi Narayan Bhandari Oct 13 '21 at 15:07