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


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

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

  • $\begingroup$ For those who are interested in the integral: here $\endgroup$ Mar 24 at 14:54
  • 1
    $\begingroup$ Thank you whoever opened this question. It was not a duplicate :) $\endgroup$
    – Kashmiri
    Mar 24 at 15:48

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.

  • $\begingroup$ As $z$--> $i$ , the real part $x$--> 0 $\endgroup$
    – Kashmiri
    Mar 24 at 10:31
  • $\begingroup$ 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 $z$). $\endgroup$
    – user
    Mar 24 at 11:30
  • $\begingroup$ Pardon me but I'm not able to understand. Could you please explain it? $\endgroup$
    – Kashmiri
    Mar 24 at 11:55
  • $\begingroup$ 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 '$x$'. $\endgroup$
    – user
    Mar 24 at 12:02
  • $\begingroup$ thanks ,but the problem wont go away even then. $\endgroup$
    – Kashmiri
    Mar 24 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.