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I want to find $\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx$

using

$$\oint_{c} \frac{e^{ix}}{1+z^2} \cdot d z$$

over the upper half of a large semicircle enclosing $z=i$ (not $\oint_{c} \frac{e^{iz}}{1+z^{2}} dz$)

For very large semicircle, the above integral reduces to $$\oint \frac{e^{i x}}{1+z^{2}} d z=\int_{-\infty}^{\infty} \frac{e^{i x}}{1+x^{2}}dx=\int_{-\infty}^{\infty} \frac{\cos x d x}{1+x^{2}}+i \int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x$$

So we have $$\oint \frac{e^{i x}}{1+z^{2}} d z=\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x+i\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x$$

Also by residue theorem $$\oint \frac{e^{i x}}{1+z^{2}} d z=2 \pi i b_{1}$$ and then

$$b_{1}=\lim _{z \rightarrow i} \frac{e^{i x}(z-i)}{1+z^{2}}=\frac{1}{2 i}$$ ( as $z$--> $i$ , the real part $x$--> $0$ )

Therefore we have$$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x+\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x=\frac{2 \pi i}{2 i}$$

And hence $$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^{2}} d x=\pi$$ which is incorrect.

Why am I getting wrong result. I suspect the above highlighted portion. Can anyone please help me. Thank you so much

Cameron Williams
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Kashmiri
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1 Answers1

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Your error is that you did not change consequently the real variable $x$ to complex variable $z$. As a consequence your integrated function is wrong as seemingly $x$ is independent of $z$.

As a result you have wrongly computed the residue at $z=i$. The correct value is: $$ \operatorname{res}(f(z),i)=\lim _{z \rightarrow i} \frac{e^{i \color{red}z}(z-i)}{1+z^{2}}=\frac{e^{-1}}{2 i}. $$

UPDATE (after OP question was edited clarifying that symbol $x$ used in the integrated function is assumed to mean the real part of $z$)

You cannot integrate the function $$f(z)=\frac{e^{i\operatorname{Re}z}}{1+z^2}$$ with the help of the residue theorem because the function is holomorphic at no point of the complex plane, whereas the residue theorem requires it to be holomorphic at any point of the part of the complex plane surrounded by the contour except for a finite set of isolated points.

To avoid this you shall instead use the meromorphic function $$ f(z)=\frac{e^{iz}}{1+z^2} $$ as suggested above.

user
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  • As $z$--> $i$ , the real part $x$--> 0 – Kashmiri Mar 24 '21 at 10:31
  • If you intend to claim that $x$ in your formulas means the real part of $z$, then you should always avoid this and express the integrated function via the variable of integration (in your case it is $z$). – user Mar 24 '21 at 11:30
  • Pardon me but I'm not able to understand. Could you please explain it? – Kashmiri Mar 24 '21 at 11:55
  • If your integrated function contains a symbol $x$ it means a parameter independent of $z$. If your intend to claim that it is a function of $z$ you should explicitly do this (e.g. replace $x$ with $\operatorname{Re} z$) or at least indicate this meaning of '$x$'. – user Mar 24 '21 at 12:02
  • thanks ,but the problem wont go away even then. – Kashmiri Mar 24 '21 at 12:49
  • See the edit of my answer. – user Mar 24 '21 at 12:50
  • Yes that's it. Basically $e^_{ix}$ isn't analytic – Kashmiri Mar 24 '21 at 14:01