1. ## Integral Equation

Hi,
I would like some help with this question :
$
\int\limits_{-\infty}^{\infty} \frac{y(\xi)d\xi }{(t-\xi)^2+25}=\frac{1}{t^2+36}
$

Find $
y(\xi)
$

This is a question from an old final exam that I'm struggling with. I think we haven't seen the necessary material this year. However, I want to be sure that I'm not missing some easy trick.
Thanks

2. Left side is a convolution product.
Use a Fourier transform on both sides.
Then isolate the Fourier transform of $y(\xi)$ by dividing on both sides, and take the inverse Fourier transform of $Y(\omega)$.
I think you should get $y(t)=\frac{1}{t^2+1}$.

3. That certainly makes sense to me now that you explained it. Thanks. However, when I back-substitute that solution into the integral equation and solve the integral numerically, I do not obtain values that are close enough to convince me it's right (I'd expect at least 2 or 3 digit agreement and preferably 6). And the numerical results reported by Mathematica do not change significantly when I up the precision of the numerical calculations. Here is an example for t=1:

Code:
In[230]:=
y[t_] := 1/(t^2 + 1);
i1 = NIntegrate[y[s]/((t - s)^2 + 25) /.
t -> 1, {s, -Infinity, Infinity}]
i2 = N[1/(t^2 + 36) /. t -> 1]

Out[231]= 0.101889

Out[232]= 0.027027
Granted, there is the chance Mathematica is not correctly doing the numerical integration. However Mathematica is usually pretty good at that especially when it does not report any complications such as highly oscillatory or very slow rate of convergence. I've tried other values of t including complex ones and the expressions do not agree.

4. You're right I typed :
Integrate[1/((1 + x^2)*((x - a)^2 + 25)), {x, -Infinity, Infinity},
Assumptions -> Im[a] == 0 ]
in mathematica and got
$

y(t)=\frac{6\pi}{5(t^2+36)}
$

$

y(t)=\frac{5}{6\pi(t^2+1)}
$

5. Very good then and thanks for that Mathematica titbit. Didn't know how to do that.

I've looked at it more carefully now and see where my problem was. This is actually a very interesting problem in contour integration as it involves the transform:

$F\left\{\frac{1}{x^2+25}\right\}=\int_{-\infty}^{\infty} \frac{e^{-isx}}{(x+5i)(x-5i)}dx$

and for now, lets not worry about the factor $\frac{1}{\sqrt{2\pi}}$ which also enters the problem up there.

Now normally, one uses a upper or lower half-disc contour and the residue at either $5i$ or $-5i$ and for the integral to be zero on the half-circle part of the contour, we need to use the appropriate residue and that depends on the sign of s and that's how the expression:

$F\left\{\frac{1}{x^2+25}\right\}=\int_{-\infty}^{\infty} \frac{e^{-isx}}{(x+5i)(x-5i)}dx=\frac{\pi}{5}e^{-5|s|}$

is obtained I think.