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I am trying to find a nicer form for the integral: $\displaystyle\int _{-\pi }^{\pi }\frac{\sin (x)}{1+x}dx$.

I considered the complex plane and I managed to get rid of the integration through an unpleasant point $x=-1$. The result is this:

$$\pi\cos(1)+\int_{0}^{+\infty}\left\{{\frac{\pi+1}{(\pi+1)^{2}+y^{2}}}+{\frac{\pi-1}{(\pi-1)^{2}+y^{2}}}\right\}\exp(-y)\;d y.$$

Is there anything else that can be done here?

Ѕᴀᴀᴅ
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Aragorn
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    If you split the integral into two pieces, perform the change of variables $y=(\pi+1)t$ and $y=(\pi-1)t$, you can use http://dlmf.nist.gov/6.7.E13 I am not sure if that helps. – Gary Nov 24 '22 at 11:12
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    Compare with this post. Perhaps also interesting is this post, for denomimator $1+x^2$ instead of $1+x$. – Dietrich Burde Nov 24 '22 at 11:12
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    https://www.wolframalpha.com/input?key=&i2d=true&i=Integrate%5BDivide%5Bsinx%2C1%2Bx%5D%2C%7Bx%2C-pi%2Cpi%7D%5D This doesn't even converge because of the asymptote at x=-1, but we can probably find its principal value using special functions. Keep in mind the following integral converges.$$\int_{-\pi}^\pi \left(\frac{\sin{x}}{1+x}+\frac{\sin(1)}{1+x}\right)$$May I ask about where this integral came from? – polychroma Nov 24 '22 at 14:29

3 Answers3

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I do not know how to evaluate this definite integral but I can help to evalaute the indefinite one of the same. $$\int\frac{\sin x}{1+x}dx$$ $$=\int\frac{\sin (u-1)}{u}du$$ where I substituted $u=x+1$ $$=\int\frac{\sin u\cos 1-\cos u\sin1}{u}du$$ $$=\cos 1\int\frac{\sin u}{u}du-\sin 1\int\frac{\cos u}{u}du$$ $$=\cos 1\cdot\textrm{Si}(u)-\sin 1\cdot\textrm{Ci}(u)+C$$ $$=\cos 1\cdot\textrm{Si}(x+1)-\sin 1\cdot\textrm{Ci}(x+1)+C$$

Vanessa
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If you use $$ \frac{1}{a^2+y^2}= \frac{1}{(y+i a)(y-ia)}=\frac i{2a}\left(\frac{1}{y+i a} -\frac{1}{y-i a}\right)$$ and with an abvious change of variable $$\int \frac{e^{-y}}{y+b}\,dy=e^b\,\, \text{Ei}(-b-y)$$ $$\int_0^\infty \frac{e^{-y}}{y+b}\,dy=e^b \,\,\Gamma (0,b)$$ you should end with $$\int_0^\infty \frac{e^{-y}}{a^2+y^2}\,dy= \frac{2 \text{Ci}(a) \sin (a)+(\pi -2 \text{Si}(a)) \cos (a)}{2 a}$$ Using it twice $$A=\pi\cos(1)+\int_{0}^{+\infty}\left\{{\frac{\pi+1}{(\pi+1)^{2}+y^{2}}}+{\frac{\pi-1}{(\pi-1)^{2}+y^{2}}}\right\}\exp(-y)\;d y$$ $$A=(\text{Ci}(-1+\pi )-\text{Ci}(1+\pi )) \sin (1)+(\text{Si}(1+\pi )-\text{Si}(1-\pi )) \cos (1)$$ which the principal value of the integral in title.

Claude Leibovici
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A physicist's approach:

Multiply the integrand by $\frac{1-x}{1-x}$ to get

$$\int _{-\pi}^{\pi}\frac{(1-x)\sin (x)}{1-x^2}dx$$

Introduce 2 parameters, $t$ and $a$

$$I(t)=\int _{-\pi}^{\pi}\frac{(1-x)\sin (xt)}{1+a^2x^2}dx$$

To return to the original integral set $t=1$ and $a=i$ where $i$ is the imaginary unit.

Take the Laplace Transform:

$$\mathcal{L} I(t)=\int _{-\pi}^{\pi}\frac{(1-x)x}{(1+a^2x^2)(s^2+x^2)}dx$$

(i will skip elementary integration and algebraic procedures)

$$\mathcal{L} I(t)=2\frac{\arctan(a\pi)-as\arctan(\frac{\pi}{s})}{a(a^2s^2-1)}$$

Now, set $a=i$ and get

$$\mathcal{L} I(t)=\int _{-\pi}^{\pi}\frac{x}{(1+x)(s^2+x^2)}dx=\frac{2s}{s^2+1}\arctan\left(\frac{\pi}{s}\right)-\frac{\ln\frac{\pi+1}{\pi-1}}{s^2+1}+\frac{\pi}{s^2+1}i$$

The imaginary part of the result is different from zero and this indicates that the integral diverges. The real part gives the Cauchy principal value (shortly $P V$)

Compactly:

$$PV\int _{-\pi}^{\pi}\frac{x}{(1+x)(s^2+x^2)}dx=\frac{2s}{s^2+1}\arctan\left(\frac{\pi}{s}\right)-\frac{\ln\frac{\pi+1}{\pi-1}}{s^2+1}$$

Now all that remains is to calculate the inverse Laplace transform of the last result. We use for this an Inverse Laplace transform table.

Final result:

$$PV\int _{-\pi}^{\pi}\frac{\sin (xt)}{1+x}dx=2\int _{0}^{t}\cos(t-\tau)\frac{\sin (\pi\tau)}{\tau}d\tau-\sin (t)\ln\frac{\pi+1}{\pi-1}$$

To get the Cauchy principal value for the original integral set $t=1$ here.

Martin Gales
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