# Thread: Complex Analysis: Residue THM problem

1. ## Complex Analysis: Residue THM problem

Q: Integral from -infinity to infinity, sinx/(x^2 + 6x + 10)

I am supposed to use residue theorem, so I tried Integral from -infinity to infinity, e^iz/(z^2 + 6z + 10)

but I get stuck after separating e^ix to cosx and i(sinx) Not sure what to do to elimiate cosx. Because of 6x, I can't say any of them are odd/even function.

Any help would be appreciated...

2. Originally Posted by hohoho00
Q: Integral from -infinity to infinity, sinx/(x^2 + 6x + 10)

I am supposed to use residue theorem, so I tried Integral from -infinity to infinity, e^iz/(z^2 + 6z + 10)

but I get stuck after separating e^ix to cosx and i(sinx) Not sure what to do to elimiate cosx. Because of 6x, I can't say any of them are odd/even function.

Any help would be appreciated...
hey mate,
Employ Euler's Identity
e^iz = cos(z) + isin(z)
thus,
e^-iz = cos(z) -isin(z)
thus,
sin(z) = (1/2i)*(e^iz + e^-iz))
(1/2i)* (e^iz + e^-iz)/(z^2 + 6z + 10) = (1/2i)( e^iz/(z^2 + 6z + 10) + e^-iz/(z^2 + 6z + 10))

obviously you have two integrals here, which unfortunately in split into 4 integrals as 1/(z^2 + 6z + 10) needs to factorised into the form 1/(z-a)(z-b) and then resolved into partial fractions, when this is achieved the Cauchy Residue Thm can be applied to evaluate the integral.

Hope this helps,

David

3. Perhaps I wasn't clear enough. I'm supposed to evaluate "Integral from -infinity to infinity, sinx/(x^2 + 6x + 10)" where x is the real value, using complex analysis residue thm.

So, letting the curve to be the top half of the circle with radius R where R --> infinity (so the curve starts from R ---> -R then from -R it crosses real axis to R), then I can separate the integral on the real axis --> hence, manipulate the function so that I can get the integrals of sinx/(x^2 + 6x + 10).

Usually, by letting e^iz on top would have been fine because I can first separate it into cosx+ isinx and then I can eliminate the odd equation (since the integral is from -infinity to infinity), but in this case, the funcion doesn't turn out to be odd, hence, cannot elimiate, which is my problem...

4. Just consider:

$\displaystyle \mathop\oint\limits_{H_u}\frac{1}{z^2+6z+10} e^{iz}dz=a+bi$

where $\displaystyle H_u$ is the upper half disc. Since you're interested in the sine integral, evaluate the contour integral and the imaginary part ($\displaystyle b$) is your integral. $\displaystyle z^2$ in the denominator assures that the integral over the half-circle contour goes to zero and we're left with:

\displaystyle \begin{aligned}\mathop\oint\limits_{H_u}\frac{1}{z ^2+6z+10} e^{iz}dz&=2\pi i\mathop\text{Res}\limits_{z=z_0}\left\{\frac{1}{z ^2+6z+10} e^{iz}\right\}\\ &=\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+6x+10}dx+i\int_{-\infty}^{\infty}\frac{\sin(x)}{x^2+6x+10}dx \end{aligned}

That is, the imaginary component of the residue calculation is the value of your integral right?

5. I get it now... since it is separated into real part and imaginary part, I just have to consider the imaginary part.

Thank you.

6. Can we assume that you have determined where the pole is and what the residue is there?

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# the residue of z/ (z-a ) z-b at infinity is

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