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Let $a\in \mathbb{C}, Im(a)\neq 0.$ Show that $\frac{2a}{\pi}\int_{0}^\infty 1/(t^2+a^2)dt$ converges and find its value.

I have idea how to approach this problem if $a$ is real number, but how can we do this one if $a\in \mathbb{C}$? I tried to write $a=x+iy, $ where $x,y\in \mathbb{R}, y\neq0$. Separating the original integral by real part and imagine part, and then trying to integral each of them by using contour integral. However, it becomes very complicated. Is there any good way to solve it? Thank you.

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    Did you mean to say that the real part of $\alpha$ is nonzero? If $\alpha=i$, for example, then the integral diverges. – 2017-02-10
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    you are right! I checked with the instructor, it shall be Real part of $a$ is nonzero. – 2017-02-25

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We first need to show the convergence of our integral. $$\int_{0}^\infty \frac 1{t^2+p^2+2ipq-q^2}dt$$
If the imaginary part of $a^2$ is zero, we get convergence iff $p^2-q^2 > 0$ which will definitely not hold because $q \in \mathbb{R}\setminus0 \implies q^2 > 0$ and to make the imaginary term go away when $q \neq 0$ we must have $p=0$.

When the imaginary part of $a^2$ is non-zero we have no poles, as $t^2$ is a real value. We thus confirm that your integral converges iff $\operatorname{Im}(a^2) \neq 0$. As far as actually solving your integral goes, things are much easier

To do so we first factor out $a^2$ and then substitute $t/a = u \implies dt=a\,du$ $$\int_{0}^\infty \frac 1{t^2+a^2}dt=\frac{1}{a^2}\int_{0}^\infty \frac 1{(t/a)^2+1}dt = \frac{1}{a}\int_{0}^\infty \frac 1{u^2+1}du$$ You should be able to take it from here I hope.

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    But wouldn't the substitution $t/a=u$ change the integration path? It is no longer the positive real axis and then it might be not apparent how the last integral is computed. – 2017-02-10
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    @MartinR If you are concerned about that then we can stop after the second step and just recognize a well known form for $\arctan(x)$, can we not? I haven't actually taken a course in Complex Analysis before and have self studied, so I could very well be wrong and would love to be corrected. – 2017-02-10
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    If $\operatorname{Re}a>0$ and $\operatorname{Im}a>0,$ then $$\int _0^\infty \frac{1}{t^2+a^2} dt=\frac{\pi}{2a}.$$ On the contrary if $\operatorname{Re}a<0$ and $\operatorname{Im}a>0$, then $$\int _0^\infty \frac{1}{t^2+a^2} dt=-\frac{\pi}{2a}.$$ – 2017-02-11
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    To ts375_zk26, Can you show me how to get this result? – 2017-02-25