How to evaluate these?!

Calculus Level 4

Let $f\left(a\right)$ be a function which satisfies

$f\left(a\right)\int_{0}^{\infty}\frac{e^{-x}}{x^{2}+a^{2}}dx=\int_{0}^{\infty}\frac{\sin x}{x+a}dx$

for all real positive $a$ . Find the value of

$f\left(\pi+e\right) f\left(\pi-e\right)$

The answer is 2.4805.

1 solution

Vilakshan Gupta
May 2, 2020

Let $I(a,k)=\int_{0}^{\infty} \dfrac{\sin x}{x+a} ~ e^{-k(x+a)} dx$

Now partially differentiating the above integral with respect to $k$ , we get $\dfrac{ \partial I}{\partial k}=- \int_{0}^{\infty} \sin x ~ e^{-k(x+a)} dx=-e^{-ka} \int_{0}^{\infty} \sin x ~ e^{-kx} dx$

Evaluating the above integral is easy by Integration by parts, which comes out to be $-e^{-ka} \int_{0}^{\infty} \sin x ~ e^{-kx} dx=- \dfrac{e^{-ak}}{1+k^2}$

And now, since $I(a,k) \to 0$ as $k \to \infty$ , we deduce that $\int_{0}^{\infty} \dfrac{\sin x}{x+a}= \lim_{k \to 0} I(a,k) = \int_{0}^{\infty} \dfrac{e^{-ka}}{k^2+1} dk = a \int_{0}^{\infty} \dfrac{e^{-u}}{u^2+a^2} du$

which implies $f(a)=a$ for $a > 0$ .

Hence $f(\pi+e)f(\pi-e)=\boxed{\pi^2-e^2 \approx 2.4805}$ .

The solution is incomprehensible near the end....Please edit it....

Aaghaz Mahajan - 1 year, 1 month ago

You mean that $\frac{\partial I}{\partial k} \; =\; -\frac{e^{-ka}}{k^2+1}$ and, since $I(a,k) \to 0$ as $k \to \infty$ , we deduce that $\int_0^\infty \frac{\sin x}{x+a}\,dx \; = \; \lim_{k \to 0}I(a,k) \; = \; \int_0^\infty \frac{e^{-ka}}{k^2+1}\,dk \; = \; a\int_0^\infty \frac{e^{-u}}{u^2+a^2}\,du$ , which makes $f(a) = a$ for $a > 0$ .

Mark Hennings - 1 year, 1 month ago

Yes sir, thanks for the correction and explanation...I have edited it.

Vilakshan Gupta - 1 year, 1 month ago

I tends to go with different approch but I got stuck with my working. Can you please help me out with this, sir ?

We note that $\frac{\sin x}{x+a}=\int_0^{\infty} e^{-(x+a)y}\sin xdy\implies \int_0^{\infty}\frac{\sin x}{x+a}dx =\int_0^{\infty}\left(\int_0^{\infty}e^{-(x+a)y}\sin xdx\right)dy$ as the $\int_0^{\infty}\left(\int_0^{\infty}e^{-(x+a)y}\sin xdy\right)dx= \int_0^{\infty}\left(\int_0^{\infty}e^{-(x+a)y}\sin xdx\right)dy$ Further it is easy to deduce by integration parts that $\int_0^{\infty}e^{-(x+a)y}\sin xdx=\left[\frac{e^{-(x+a)y}(y\sin x+\cos x)}{1+y^2}\right]_{0}^{\infty}=\frac{e^{-ay}}{1+y^2}$ and hence $\int_0^{\infty}\frac{\sin x}{x+a}dx= \int_0^{\infty}\left(\int_0^{\infty}e^{-(x+a)y}\sin xdx\right)dy=\int_0^{\infty}\frac{e^{-ay}}{1+y^2}dy$ The latter integral is in the form of Exponential integral but I don't know what to do after this as setting $u=1+y^2$ doesn't work.

Added : I even tried up doing $y^2+1=(y+i)(y-i)$ and got the indefinite integral in terms of exponential integral ( where $i^2=-1$ ) but I don't know how closed form it appears in sine and cosine integral( as per WA).

Update: I solved it sir.

Naren Bhandari - 1 year, 1 month ago

How to evaluate these?!

The answer is 2.4805.

1 solution

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